Chapter 7 Fractions

A fraction represents a part of a whole. It is written as $\frac{\text{Numerator}}{\text{Denominator}}$, where the denominator represents the total number of equal parts the whole is divided into, and the numerator represents the number of parts being considered (in this case, the shaded parts).

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(i)

The figure is divided into 4 equal parts.

Number of shaded parts = 2.

Fraction = $\frac{2}{4}$.

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(ii)

The figure is divided into 9 equal parts.

Number of shaded parts = 8.

Fraction = $\frac{8}{9}$.

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(iii)

There are a total of 8 balls.

Number of shaded balls = 4.

Fraction = $\frac{4}{8}$.

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(iv)

The figure is divided into 4 equal parts.

Number of shaded parts = 1.

Fraction = $\frac{1}{4}$.

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(v)

The figure is divided into 7 equal parts.

Number of shaded parts = 3.

Fraction = $\frac{3}{7}$.

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(vi)

There are a total of 12 flowers.

Number of shaded flowers = 3.

Fraction = $\frac{3}{12}$.

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(vii)

There are a total of 10 pencils.

Number of shaded pencils = 10.

Fraction = $\frac{10}{10}$.

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(viii)

The figure is divided into 9 equal parts.

Number of shaded parts = 4.

Fraction = $\frac{4}{9}$.

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(ix)

The figure is divided into 8 equal parts.

Number of shaded parts = 4.

Fraction = $\frac{4}{8}$.

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(x)

The figure is divided into 2 equal parts.

Number of shaded parts = 1.

Fraction = $\frac{1}{2}$.

To color the part according to the given fraction, we look at the denominator for the total number of equal parts and the numerator for the number of parts to be colored.

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(i) Fraction: $\frac{1}{6}$

The circle is divided into 6 equal parts. The numerator is 1, so we need to color 1 of these parts.

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(ii) Fraction: $\frac{1}{4}$

The triangle is divided into 4 equal parts. The numerator is 1, so we need to color 1 of these parts.

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(iii) Fraction: $\frac{1}{3}$

The rectangle is divided into 3 equal parts. The numerator is 1, so we need to color 1 of these parts.

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(iv) Fraction: $\frac{3}{4}$

The figure contains 4 equal parts (balls). The numerator is 3, so we need to color 3 of these parts.

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(v) Fraction: $\frac{4}{9}$

The rectangle is divided into 9 equal squares (a 3x3 grid). The numerator is 4, so we need to color 4 of these squares.

Solution:

To represent a fraction as a shaded portion of a figure, the fundamental condition is that the figure must be divided into equal parts. The denominator represents the total number of equal parts, and the numerator represents the number of those equal parts that are shaded.

Let's examine each figure:

Figure 1 (Triangle):

The figure claims to represent the fraction $\frac{1}{2}$.

This means the triangle should be divided into 2 equal parts, and 1 part should be shaded.

However, the figure shows the triangle divided into two parts that are clearly not equal in area.

Error: The parts are not equal.

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Figure 2 (Rectangle):

The figure claims to represent the fraction $\frac{1}{4}$.

This means the rectangle should be divided into 4 equal parts, and 1 part should be shaded.

However, the figure shows the rectangle divided into four parts that are not equal in area.

Error: The parts are not equal.

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Figure 3 (Circle):

The figure claims to represent the fraction $\frac{3}{4}$.

This means the circle should be divided into 4 equal parts, and 3 parts should be shaded.

However, the figure shows the circle divided into four parts that are not equal in area.

Error: The parts are not equal.

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In conclusion, the error in all three figures is that the whole figure is not divided into equal parts, which is a necessary condition for representing the shaded portion as a fraction.

Solution:

To find the fraction of a day that represents 8 hours, we need to express both quantities in the same unit.

We know that 1 day is equivalent to 24 hours.

The total number of hours in a day (the whole) is 24 hours.

The given number of hours (the part) is 8 hours.

The required fraction is formed by placing the part over the whole:

Fraction = $\frac{\text{Given number of hours}}{\text{Total number of hours in a day}}$

Fraction = $\frac{8}{24}$

Now, we simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 8:

Fraction = $\frac{8 \div 8}{24 \div 8} = \frac{1}{3}$

Therefore, 8 hours is $\mathbf{\frac{1}{3}}$ of a day.

Solution:

To find the fraction of an hour that represents 40 minutes, we need to express both quantities in the same unit, which is minutes in this case.

We know that 1 hour is equivalent to 60 minutes.

The total number of minutes in an hour (the whole) is 60 minutes.

The given number of minutes (the part) is 40 minutes.

The required fraction is formed by placing the part over the whole:

Fraction = $\frac{\text{Given number of minutes}}{\text{Total number of minutes in an hour}}$

Fraction = $\frac{40}{60}$

Now, we simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor. The GCD of 40 and 60 is 20.

Fraction = $\frac{40 \div 20}{60 \div 20} = \frac{2}{3}$

Alternatively, we can simplify by cancelling common factors:

Fraction = $\frac{\cancel{40}^{2}}{\cancel{60}_{3}} = \frac{2}{3}$

Therefore, 40 minutes is $\mathbf{\frac{2}{3}}$ of an hour.

Solution:

There are three boys in total: Arya, Abhimanyu, and Vivek.

Arya has two sandwiches: one vegetable sandwich and one jam sandwich.

The goal is to share these two sandwiches equally among the three boys.

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(a) How Arya can divide his sandwiches:

To ensure each person gets an equal share of each sandwich, Arya needs to divide each sandwich into the same number of equal parts as there are people sharing.

Number of people = 3.

So, Arya must divide each sandwich into 3 equal parts.

Specifically:

1. Divide the vegetable sandwich into 3 equal parts.

2. Divide the jam sandwich into 3 equal parts.

Then, each boy (Arya, Abhimanyu, and Vivek) will take one part from the vegetable sandwich and one part from the jam sandwich.

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(b) What part of a sandwich each boy will receive:

Consider one sandwich (either vegetable or jam). It is divided into 3 equal parts because there are 3 boys sharing.

Each boy receives 1 of these equal parts from each sandwich.

The fraction representing the part of a sandwich received by each boy is:

Fraction = $\frac{\text{Number of parts received by one boy}}{\text{Total number of equal parts the sandwich is divided into}}$

Fraction = $\frac{1}{3}$

Therefore, each boy will receive $\mathbf{\frac{1}{3}}$ of a sandwich.

Solution:

To find the fraction of dresses Kanchan has finished, we need to compare the number of dresses finished to the total number of dresses she had to dye.

Total number of dresses Kanchan had to dye (the whole) = 30.

Number of dresses Kanchan has finished so far (the part) = 20.

The fraction of dresses finished is given by:

Fraction = $\frac{\text{Number of dresses finished}}{\text{Total number of dresses}}$

Fraction = $\frac{20}{30}$

Now, we simplify the fraction to its lowest terms. We can divide both the numerator and the denominator by their greatest common divisor, which is 10.

Fraction = $\frac{20 \div 10}{30 \div 10} = \frac{2}{3}$

Alternatively, using cancellation:

Fraction = $\frac{\cancel{20}^{2}}{\cancel{30}_{3}} = \frac{2}{3}$

Therefore, Kanchan has finished $\mathbf{\frac{2}{3}}$ of the dresses.

Solution:

First, let's list the natural numbers from 2 to 12:

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Next, let's count the total number of these natural numbers.

Total number of natural numbers from 2 to 12 = 11.

Now, we need to identify the prime numbers within this list. A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself.

Let's check each number:

- 2 is a prime number (divisors: 1, 2).

- 3 is a prime number (divisors: 1, 3).

- 4 is not a prime number (divisors: 1, 2, 4).

- 5 is a prime number (divisors: 1, 5).

- 6 is not a prime number (divisors: 1, 2, 3, 6).

- 7 is a prime number (divisors: 1, 7).

- 8 is not a prime number (divisors: 1, 2, 4, 8).

- 9 is not a prime number (divisors: 1, 3, 9).

- 10 is not a prime number (divisors: 1, 2, 5, 10).

- 11 is a prime number (divisors: 1, 11).

- 12 is not a prime number (divisors: 1, 2, 3, 4, 6, 12).

The prime numbers in the list are: 2, 3, 5, 7, 11.

Let's count the number of prime numbers identified.

Number of prime numbers = 5.

The fraction of prime numbers is the ratio of the number of prime numbers to the total number of natural numbers from 2 to 12.

Fraction = $\frac{\text{Number of prime numbers}}{\text{Total number of natural numbers from 2 to 12}}$

Fraction = $\frac{5}{11}$

Therefore, the fraction of prime numbers among the natural numbers from 2 to 12 is $\mathbf{\frac{5}{11}}$.

Solution:

First, let's list the natural numbers from 102 to 113:

102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113

Next, let's count the total number of these natural numbers.

Total number of natural numbers from 102 to 113 = 12.

Now, we need to identify the prime numbers within this list. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's check each number:

- 102 is not prime (it's even, divisible by 2).

- 103 is a prime number (divisors: 1, 103).

- 104 is not prime (it's even, divisible by 2).

- 105 is not prime (ends in 5, divisible by 5).

- 106 is not prime (it's even, divisible by 2).

- 107 is a prime number (divisors: 1, 107).

- 108 is not prime (it's even, divisible by 2).

- 109 is a prime number (divisors: 1, 109).

- 110 is not prime (ends in 0, divisible by 10).

- 111 is not prime (sum of digits $1+1+1=3$, divisible by 3).

- 112 is not prime (it's even, divisible by 2).

- 113 is a prime number (divisors: 1, 113).

The prime numbers in the list are: 103, 107, 109, 113.

Let's count the number of prime numbers identified.

Number of prime numbers = 4.

The fraction of prime numbers is the ratio of the number of prime numbers to the total number of natural numbers from 102 to 113.

Fraction = $\frac{\text{Number of prime numbers}}{\text{Total number of natural numbers from 102 to 113}}$

Fraction = $\frac{4}{12}$

Now, we simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 4:

Fraction = $\frac{4 \div 4}{12 \div 4} = \frac{1}{3}$

Alternatively, using cancellation:

Fraction = $\frac{\cancel{4}^{1}}{\cancel{12}_{3}} = \frac{1}{3}$

Therefore, the fraction of prime numbers among the natural numbers from 102 to 113 is $\mathbf{\frac{1}{3}}$.

Solution:

First, we need to determine the total number of circles shown in the image.

By counting, we find there are 8 circles in total.

Next, we need to count the number of circles that have an 'X' inside them.

By counting, we find there are 4 circles with an 'X' in them.

The fraction of circles with an 'X' is the ratio of the number of circles with an 'X' to the total number of circles.

Fraction = $\frac{\text{Number of circles with an 'X'}}{\text{Total number of circles}}$

Fraction = $\frac{4}{8}$

Now, we simplify this fraction to its lowest terms by dividing both the numerator and the denominator by their greatest common divisor, which is 4.

Fraction = $\frac{4 \div 4}{8 \div 4} = \frac{1}{2}$

Alternatively, using cancellation:

Fraction = $\frac{\cancel{4}^{1}}{\cancel{8}_{2}} = \frac{1}{2}$

Therefore, $\mathbf{\frac{1}{2}}$ of the circles have X's in them.

Solution:

First, let's determine the total number of CDs Kristin has.

Number of CDs Kristin bought = 3.

Number of CDs Kristin received as gifts = 5.

Total number of CDs = Number of CDs bought + Number of CDs received as gifts

Total number of CDs = $3 + 5 = 8$.

So, Kristin has a total of 8 CDs.

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Now, let's find the fraction of her total CDs that she bought.

Fraction bought = $\frac{\text{Number of CDs bought}}{\text{Total number of CDs}}$

Fraction bought = $\frac{3}{8}$

This fraction is already in its simplest form.

Therefore, the fraction of CDs Kristin bought is $\mathbf{\frac{3}{8}}$.

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Next, let's find the fraction of her total CDs that she received as gifts.

Fraction received as gifts = $\frac{\text{Number of CDs received as gifts}}{\text{Total number of CDs}}$

Fraction received as gifts = $\frac{5}{8}$

This fraction is also in its simplest form.

Therefore, the fraction of CDs Kristin received as gifts is $\mathbf{\frac{5}{8}}$.

Solution:

To convert an improper fraction (where the numerator is greater than or equal to the denominator) to a mixed fraction, we divide the numerator by the denominator. The quotient becomes the whole number part, the remainder becomes the numerator of the fractional part, and the denominator stays the same.

Mixed Fraction = Quotient $\frac{\text{Remainder}}{\text{Denominator}}$

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(a) $\frac{17}{4}$

Divide 17 by 4:

$\begin{array}{r} 4\phantom{)} \\ 4{\overline{\smash{\big)}\,17\phantom{)}}} \\ \underline{-~\phantom{(}16} \\ 1\phantom{)} \end{array}$

Quotient = 4

Remainder = 1

Denominator = 4

So, the mixed fraction is $\mathbf{4\frac{1}{4}}$.

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(b) $\frac{11}{3}$

Divide 11 by 3:

$\begin{array}{r} 3\phantom{)} \\ 3{\overline{\smash{\big)}\,11\phantom{)}}} \\ \underline{-~\phantom{(}9} \\ 2\phantom{)} \end{array}$

Quotient = 3

Remainder = 2

Denominator = 3

So, the mixed fraction is $\mathbf{3\frac{2}{3}}$.

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(c) $\frac{27}{5}$

Divide 27 by 5:

$\begin{array}{r} 5\phantom{)} \\ 5{\overline{\smash{\big)}\,27\phantom{)}}} \\ \underline{-~\phantom{(}25} \\ 2\phantom{)} \end{array}$

Quotient = 5

Remainder = 2

Denominator = 5

So, the mixed fraction is $\mathbf{5\frac{2}{5}}$.

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(d) $\frac{7}{3}$

Divide 7 by 3:

$\begin{array}{r} 2\phantom{)} \\ 3{\overline{\smash{\big)}\,7\phantom{)}}} \\ \underline{-~\phantom{(}6} \\ 1\phantom{)} \end{array}$

Quotient = 2

Remainder = 1

Denominator = 3

So, the mixed fraction is $\mathbf{2\frac{1}{3}}$.

Solution:

To convert a mixed fraction to an improper fraction, we use the following formula:

Improper Fraction = $\frac{(\text{Whole number} \times \text{Denominator}) + \text{Numerator}}{\text{Denominator}}$

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(a) $2\frac{3}{4}$

Here, Whole number = 2, Numerator = 3, Denominator = 4.

Improper Fraction = $\frac{(2 \times 4) + 3}{4}$

Improper Fraction = $\frac{8 + 3}{4}$

Improper Fraction = $\mathbf{\frac{11}{4}}$

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(b) $7\frac{1}{9}$

Here, Whole number = 7, Numerator = 1, Denominator = 9.

Improper Fraction = $\frac{(7 \times 9) + 1}{9}$

Improper Fraction = $\frac{63 + 1}{9}$

Improper Fraction = $\mathbf{\frac{64}{9}}$

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(c) $5\frac{3}{7}$

Here, Whole number = 5, Numerator = 3, Denominator = 7.

Improper Fraction = $\frac{(5 \times 7) + 3}{7}$

Improper Fraction = $\frac{35 + 3}{7}$

Improper Fraction = $\mathbf{\frac{38}{7}}$

We will represent each set of fractions on a separate number line.

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(a) $\frac{1}{2}$ , $\frac{1}{4}$ , $\frac{3}{4}$ , $\frac{4}{4}$

To plot these fractions, we can use a common denominator, which is 4. The fraction $\frac{1}{2}$ is equivalent to $\frac{2}{4}$. We need to divide the segment between 0 and 1 into 4 equal parts.

The points are located as follows:

• The first mark after 0 is $\frac{1}{4}$.
• The second mark is $\frac{2}{4}$, which is $\frac{1}{2}$.
• The third mark is $\frac{3}{4}$.
• The fourth mark is $\frac{4}{4}$, which is 1.

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(b) $\frac{1}{8}$ , $\frac{2}{8}$ , $\frac{3}{8}$ , $\frac{7}{8}$

For these fractions, the denominator is 8. We need to divide the segment between 0 and 1 into 8 equal parts.

The points are located as follows:

• The first mark after 0 is $\frac{1}{8}$.
• The second mark is $\frac{2}{8}$.
• The third mark is $\frac{3}{8}$.
• The seventh mark is $\frac{7}{8}$.

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(c) $\frac{2}{5}$ , $\frac{3}{5}$ , $\frac{8}{5}$ , $\frac{4}{5}$

For these fractions, the denominator is 5. Since we have an improper fraction $\frac{8}{5}$ ($1\frac{3}{5}$), the number line must extend beyond 1. We will divide each unit segment (from 0 to 1, and 1 to 2) into 5 equal parts.

The points are located as follows:

• The second mark after 0 is $\frac{2}{5}$.
• The third mark after 0 is $\frac{3}{5}$.
• The fourth mark after 0 is $\frac{4}{5}$.
• The eighth mark after 0 (or the third mark after 1) is $\frac{8}{5}$.

Solution:

To convert an improper fraction (where the numerator is greater than or equal to the denominator) into a mixed fraction, we perform division. Divide the numerator by the denominator to get a quotient and a remainder.

The mixed fraction is then represented as: $\text{Quotient} \frac{\text{Remainder}}{\text{Denominator}}$.

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(a) $\frac{20}{3}$

Divide 20 by 3:

$\begin{array}{r} 6\phantom{)} \\ 3{\overline{\smash{\big)}\,20\phantom{)}}} \\ \underline{-~\phantom{(}18} \\ 2\phantom{)} \end{array}$

Quotient = 6, Remainder = 2, Denominator = 3.

Therefore, the mixed fraction is $\mathbf{6\frac{2}{3}}$.

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(b) $\frac{11}{5}$

Divide 11 by 5:

$\begin{array}{r} 2\phantom{)} \\ 5{\overline{\smash{\big)}\,11\phantom{)}}} \\ \underline{-~\phantom{(}10} \\ 1\phantom{)} \end{array}$

Quotient = 2, Remainder = 1, Denominator = 5.

Therefore, the mixed fraction is $\mathbf{2\frac{1}{5}}$.

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(c) $\frac{17}{7}$

Divide 17 by 7:

$\begin{array}{r} 2\phantom{)} \\ 7{\overline{\smash{\big)}\,17\phantom{)}}} \\ \underline{-~\phantom{(}14} \\ 3\phantom{)} \end{array}$

Quotient = 2, Remainder = 3, Denominator = 7.

Therefore, the mixed fraction is $\mathbf{2\frac{3}{7}}$.

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(d) $\frac{28}{5}$

Divide 28 by 5:

$\begin{array}{r} 5\phantom{)} \\ 5{\overline{\smash{\big)}\,28\phantom{)}}} \\ \underline{-~\phantom{(}25} \\ 3\phantom{)} \end{array}$

Quotient = 5, Remainder = 3, Denominator = 5.

Therefore, the mixed fraction is $\mathbf{5\frac{3}{5}}$.

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(e) $\frac{19}{6}$

Divide 19 by 6:

$\begin{array}{r} 3\phantom{)} \\ 6{\overline{\smash{\big)}\,19\phantom{)}}} \\ \underline{-~\phantom{(}18} \\ 1\phantom{)} \end{array}$

Quotient = 3, Remainder = 1, Denominator = 6.

Therefore, the mixed fraction is $\mathbf{3\frac{1}{6}}$.

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(f) $\frac{35}{9}$

Divide 35 by 9:

$\begin{array}{r} 3\phantom{)} \\ 9{\overline{\smash{\big)}\,35\phantom{)}}} \\ \underline{-~\phantom{(}27} \\ 8\phantom{)} \end{array}$

Quotient = 3, Remainder = 8, Denominator = 9.

Therefore, the mixed fraction is $\mathbf{3\frac{8}{9}}$.

Solution:

To convert a mixed fraction into an improper fraction, we multiply the whole number part by the denominator of the fractional part and add the numerator of the fractional part. This result becomes the new numerator, and the denominator remains the same.

Formula: Improper Fraction = $\frac{(\text{Whole number} \times \text{Denominator}) + \text{Numerator}}{\text{Denominator}}$

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(a) $7\frac{3}{4}$

Here, Whole number = 7, Numerator = 3, Denominator = 4.

Improper Fraction = $\frac{(7 \times 4) + 3}{4}$

Improper Fraction = $\frac{28 + 3}{4}$

Improper Fraction = $\mathbf{\frac{31}{4}}$

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(b) $5\frac{6}{7}$

Here, Whole number = 5, Numerator = 6, Denominator = 7.

Improper Fraction = $\frac{(5 \times 7) + 6}{7}$

Improper Fraction = $\frac{35 + 6}{7}$

Improper Fraction = $\mathbf{\frac{41}{7}}$

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(c) $2\frac{5}{6}$

Here, Whole number = 2, Numerator = 5, Denominator = 6.

Improper Fraction = $\frac{(2 \times 6) + 5}{6}$

Improper Fraction = $\frac{12 + 5}{6}$

Improper Fraction = $\mathbf{\frac{17}{6}}$

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(d) $10\frac{3}{5}$

Here, Whole number = 10, Numerator = 3, Denominator = 5.

Improper Fraction = $\frac{(10 \times 5) + 3}{5}$

Improper Fraction = $\frac{50 + 3}{5}$

Improper Fraction = $\mathbf{\frac{53}{5}}$

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(e) $9\frac{3}{7}$

Here, Whole number = 9, Numerator = 3, Denominator = 7.

Improper Fraction = $\frac{(9 \times 7) + 3}{7}$

Improper Fraction = $\frac{63 + 3}{7}$

Improper Fraction = $\mathbf{\frac{66}{7}}$

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(f) $8\frac{4}{9}$

Here, Whole number = 8, Numerator = 4, Denominator = 9.

Improper Fraction = $\frac{(8 \times 9) + 4}{9}$

Improper Fraction = $\frac{72 + 4}{9}$

Improper Fraction = $\mathbf{\frac{76}{9}}$

Solution:

We are given the fraction $\frac{2}{5}$ and we want to find an equivalent fraction whose numerator is 6.

Equivalent fractions represent the same value, and they can be obtained by multiplying or dividing both the numerator and the denominator of a fraction by the same non-zero number.

Let the equivalent fraction be $\frac{6}{x}$, where $x$ is the unknown denominator.

So, we have $\frac{2}{5} = \frac{6}{x}$.

We need to find the number by which the original numerator (2) must be multiplied to get the new numerator (6).

This number is $6 \div 2 = 3$.

To keep the fraction equivalent, we must multiply the original denominator (5) by the same number (3).

New denominator = $5 \times 3 = 15$.

Therefore, the equivalent fraction is $\mathbf{\frac{6}{15}}$.

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Alternate Method:

We are looking for a number $k$ such that $\frac{2 \times k}{5 \times k} = \frac{6}{\text{New Denominator}}$.

From the numerator, we have $2 \times k = 6$.

Solving for $k$, we get $k = \frac{6}{2} = 3$.

Now, multiply the denominator by this factor $k=3$.

New Denominator = $5 \times k = 5 \times 3 = 15$.

Thus, the equivalent fraction is $\mathbf{\frac{6}{15}}$.

Solution:

We are given the fraction $\frac{15}{35}$ and we want to find an equivalent fraction whose denominator is 7.

Equivalent fractions represent the same value. They can be obtained by multiplying or dividing both the numerator and the denominator of a fraction by the same non-zero number.

Let the equivalent fraction be $\frac{x}{7}$, where $x$ is the unknown numerator.

So, we have $\frac{15}{35} = \frac{x}{7}$.

We need to find the number by which the original denominator (35) must be divided to get the new denominator (7).

This number is $35 \div 7 = 5$.

To keep the fraction equivalent, we must divide the original numerator (15) by the same number (5).

New numerator = $15 \div 5 = 3$.

Therefore, the equivalent fraction is $\mathbf{\frac{3}{7}}$.

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Alternate Method:

We are looking for a number $k$ such that $\frac{15 \div k}{35 \div k} = \frac{\text{New Numerator}}{7}$.

From the denominator, we have $35 \div k = 7$.

Solving for $k$, we get $k = \frac{35}{7} = 5$.

Now, divide the original numerator by this factor $k=5$.

New Numerator = $15 \div k = 15 \div 5 = 3$.

Thus, the equivalent fraction is $\mathbf{\frac{3}{7}}$.

Solution:

We are given the fraction $\frac{2}{9}$ and we need to find an equivalent fraction that has a denominator of 63.

Equivalent fractions are obtained by multiplying or dividing both the numerator and the denominator of a given fraction by the same non-zero number.

Let the required equivalent fraction be $\frac{x}{63}$, where $x$ is the unknown numerator.

We have the relationship: $\frac{2}{9} = \frac{x}{63}$.

To find the new numerator $x$, we first determine the factor by which the original denominator (9) was multiplied to get the new denominator (63).

Factor = $\frac{\text{New Denominator}}{\text{Original Denominator}} = \frac{63}{9} = 7$.

Since the denominator was multiplied by 7, we must multiply the original numerator (2) by the same factor (7) to get the new numerator $x$.

New Numerator $x = \text{Original Numerator} \times \text{Factor} = 2 \times 7 = 14$.

Therefore, the equivalent fraction of $\frac{2}{9}$ with denominator 63 is $\mathbf{\frac{14}{63}}$.

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Alternate Method:

Let the equivalent fraction be $\frac{2 \times k}{9 \times k}$ for some non-zero integer $k$.

We are given that the new denominator is 63.

So, $9 \times k = 63$.

Solving for $k$, we get $k = \frac{63}{9} = 7$.

Now, find the new numerator by multiplying the original numerator by $k=7$.

New Numerator = $2 \times k = 2 \times 7 = 14$.

Thus, the equivalent fraction is $\mathbf{\frac{14}{63}}$.

We need to determine the fraction representing the shaded portion for each figure in both sets (a) and (b), and then check if the fractions within each set are equivalent.

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(a)

(i) The figure is divided into 2 equal parts, with 1 part shaded. The fraction is $\frac{1}{2}$.

(ii) The figure is divided into 4 equal parts, with 2 parts shaded. The fraction is $\frac{2}{4}$. Simplifying this gives $\frac{\cancel{2}^1}{\cancel{4}_2} = \mathbf{\frac{1}{2}}$.

(iii) The figure is divided into 6 equal parts, with 3 parts shaded. The fraction is $\frac{3}{6}$. Simplifying this gives $\frac{\cancel{3}^1}{\cancel{6}_2} = \mathbf{\frac{1}{2}}$.

(iv) The figure is divided into 8 equal parts, with 4 parts shaded. The fraction is $\frac{4}{8}$. Simplifying this gives $\frac{\cancel{4}^1}{\cancel{8}_2} = \mathbf{\frac{1}{2}}$.

Conclusion for (a): Since all the fractions simplify to the same value ($\frac{1}{2}$), yes, all these fractions are equivalent.

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(b)

(i) The figure has 12 equal parts, with 4 parts shaded. The fraction is $\frac{4}{12}$. Simplifying this gives $\frac{\cancel{4}^1}{\cancel{12}_3} = \mathbf{\frac{1}{3}}$.

(ii) The figure has 9 equal parts, with 3 parts shaded. The fraction is $\frac{3}{9}$. Simplifying this gives $\frac{\cancel{3}^1}{\cancel{9}_3} = \mathbf{\frac{1}{3}}$.

(iii) The figure has 6 equal parts, with 2 parts shaded. The fraction is $\frac{2}{6}$. Simplifying this gives $\frac{\cancel{2}^1}{\cancel{6}_3} = \mathbf{\frac{1}{3}}$.

(iv) The figure has 3 equal parts, with 1 part shaded. The fraction is $\frac{1}{3}$.

(v) The figure has 15 equal parts, with 6 parts shaded. The fraction is $\frac{6}{15}$. Simplifying this by dividing both numerator and denominator by 3 gives $\frac{\cancel{6}^2}{\cancel{15}_5} = \mathbf{\frac{2}{5}}$.

Conclusion for (b): The simplified fractions are $\frac{1}{3}$, $\frac{1}{3}$, $\frac{1}{3}$, $\frac{1}{3}$, and $\frac{2}{5}$. Since the simplified form of all the fractions is not the same, no, these fractions are not equivalent.

Solution:

First, let's write the fraction represented by the shaded portion for each figure in both rows and simplify them.

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Row 1 Fractions:

(a) Total parts = 2, Shaded parts = 1. Fraction = $\mathbf{\frac{1}{2}}$. (Already simplified)

(b) Total parts = 6, Shaded parts = 4. Fraction = $\frac{4}{6}$. Simplified = $\frac{4 \div 2}{6 \div 2} = \mathbf{\frac{2}{3}}$.

(c) Total parts = 9, Shaded parts = 3. Fraction = $\frac{3}{9}$. Simplified = $\frac{3 \div 3}{9 \div 3} = \mathbf{\frac{1}{3}}$.

(d) Total parts = 8, Shaded parts = 2. Fraction = $\frac{2}{8}$. Simplified = $\frac{2 \div 2}{8 \div 2} = \mathbf{\frac{1}{4}}$.

(e) Total parts = 4, Shaded parts = 3. Fraction = $\mathbf{\frac{3}{4}}$. (Already simplified)

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Row 2 Fractions:

(i) Total parts = 18, Shaded parts = 6. Fraction = $\frac{6}{18}$. Simplified = $\frac{6 \div 6}{18 \div 6} = \mathbf{\frac{1}{3}}$.

(ii) Total parts = 8, Shaded parts = 4. Fraction = $\frac{4}{8}$. Simplified = $\frac{4 \div 4}{8 \div 4} = \mathbf{\frac{1}{2}}$.

(iii) Total parts = 16, Shaded parts = 12. Fraction = $\frac{12}{16}$. Simplified = $\frac{12 \div 4}{16 \div 4} = \mathbf{\frac{3}{4}}$.

(iv) Total parts = 12, Shaded parts = 8. Fraction = $\frac{8}{12}$. Simplified = $\frac{8 \div 4}{12 \div 4} = \mathbf{\frac{2}{3}}$.

(v) Total parts = 16, Shaded parts = 4. Fraction = $\frac{4}{16}$. Simplified = $\frac{4 \div 4}{16 \div 4} = \mathbf{\frac{1}{4}}$.

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Pairing Equivalent Fractions:

Now, we match the simplified fractions from Row 1 with those from Row 2.

(a) $\frac{1}{2}$ is equivalent to (ii) $\frac{1}{2}$. $\quad \rightarrow \quad$ (a) and (ii)

(b) $\frac{2}{3}$ is equivalent to (iv) $\frac{2}{3}$. $\quad \rightarrow \quad$ (b) and (iv)

(c) $\frac{1}{3}$ is equivalent to (i) $\frac{1}{3}$. $\quad \rightarrow \quad$ (c) and (i)

(d) $\frac{1}{4}$ is equivalent to (v) $\frac{1}{4}$. $\quad \rightarrow \quad$ (d) and (v)

(e) $\frac{3}{4}$ is equivalent to (iii) $\frac{3}{4}$. $\quad \rightarrow \quad$ (e) and (iii)

Solution:

To find the missing number in equivalent fractions, we determine the factor by which the known part (numerator or denominator) was multiplied or divided, and then apply the same operation to the other part.

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(a) $\frac{2}{7} = \frac{8}{\boxed{?}}$

The numerator changes from 2 to 8. We ask: $2 \times ? = 8$. The factor is $8 \div 2 = 4$.

To maintain equivalence, we multiply the denominator by the same factor (4).

Missing denominator = $7 \times 4 = 28$.

So, $\frac{2}{7} = \frac{8}{\mathbf{28}}$. The correct number is 28.

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(b) $\frac{5}{8} = \frac{10}{\boxed{?}}$

The numerator changes from 5 to 10. The factor is $10 \div 5 = 2$.

Multiply the denominator by the same factor (2).

Missing denominator = $8 \times 2 = 16$.

So, $\frac{5}{8} = \frac{10}{\mathbf{16}}$. The correct number is 16.

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(c) $\frac{3}{5} = \frac{\boxed{?}}{20}$

The denominator changes from 5 to 20. The factor is $20 \div 5 = 4$.

Multiply the numerator by the same factor (4).

Missing numerator = $3 \times 4 = 12$.

So, $\frac{3}{5} = \frac{\mathbf{12}}{20}$. The correct number is 12.

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(d) $\frac{45}{60} = \frac{15}{\boxed{?}}$

The numerator changes from 45 to 15. We ask: $45 \div ? = 15$. The factor is $45 \div 15 = 3$.

To maintain equivalence, we divide the denominator by the same factor (3).

Missing denominator = $60 \div 3 = 20$.

So, $\frac{45}{60} = \frac{15}{\mathbf{20}}$. The correct number is 20.

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(e) $\frac{18}{24} = \frac{\boxed{?}}{4}$

The denominator changes from 24 to 4. The factor is $24 \div 4 = 6$.

Divide the numerator by the same factor (6).

Missing numerator = $18 \div 6 = 3$.

So, $\frac{18}{24} = \frac{\mathbf{3}}{4}$. The correct number is 3.

Solution:

We start with the fraction $\frac{3}{5}$. To find an equivalent fraction with a given numerator or denominator, we determine the factor by which the original part was multiplied or divided and apply the same operation to the other part.

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(a) Denominator 20

We want to find an equivalent fraction $\frac{x}{20}$.

Compare the denominators: The original denominator is 5, and the new denominator is 20.

The factor by which the denominator was multiplied is $20 \div 5 = 4$.

Multiply the original numerator (3) by the same factor (4):

New numerator $x = 3 \times 4 = 12$.

Therefore, the equivalent fraction is $\mathbf{\frac{12}{20}}$.

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(b) Numerator 9

We want to find an equivalent fraction $\frac{9}{y}$.

Compare the numerators: The original numerator is 3, and the new numerator is 9.

The factor by which the numerator was multiplied is $9 \div 3 = 3$.

Multiply the original denominator (5) by the same factor (3):

New denominator $y = 5 \times 3 = 15$.

Therefore, the equivalent fraction is $\mathbf{\frac{9}{15}}$.

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(c) Denominator 30

We want to find an equivalent fraction $\frac{x}{30}$.

Compare the denominators: The original denominator is 5, and the new denominator is 30.

The factor by which the denominator was multiplied is $30 \div 5 = 6$.

Multiply the original numerator (3) by the same factor (6):

New numerator $x = 3 \times 6 = 18$.

Therefore, the equivalent fraction is $\mathbf{\frac{18}{30}}$.

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(d) Numerator 27

We want to find an equivalent fraction $\frac{27}{y}$.

Compare the numerators: The original numerator is 3, and the new numerator is 27.

The factor by which the numerator was multiplied is $27 \div 3 = 9$.

Multiply the original denominator (5) by the same factor (9):

New denominator $y = 5 \times 9 = 45$.

Therefore, the equivalent fraction is $\mathbf{\frac{27}{45}}$.

Solution:

We start with the fraction $\frac{36}{48}$. To find an equivalent fraction with a smaller numerator or denominator, we need to divide both the numerator and the denominator by the same non-zero number (a common factor).

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(a) Numerator 9

We want to find an equivalent fraction $\frac{9}{x}$.

Compare the numerators: The original numerator is 36, and the new numerator is 9.

To get from 36 to 9, we need to divide by a certain factor.

Factor = $\frac{\text{Original Numerator}}{\text{New Numerator}} = \frac{36}{9} = 4$.

To find the equivalent fraction, we must divide the original denominator (48) by the same factor (4).

New denominator $x = \frac{\text{Original Denominator}}{\text{Factor}} = \frac{48}{4} = 12$.

Therefore, the equivalent fraction is $\mathbf{\frac{9}{12}}$.

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(b) Denominator 4

We want to find an equivalent fraction $\frac{y}{4}$.

Compare the denominators: The original denominator is 48, and the new denominator is 4.

To get from 48 to 4, we need to divide by a certain factor.

Factor = $\frac{\text{Original Denominator}}{\text{New Denominator}} = \frac{48}{4} = 12$.

To find the equivalent fraction, we must divide the original numerator (36) by the same factor (12).

New numerator $y = \frac{\text{Original Numerator}}{\text{Factor}} = \frac{36}{12} = 3$.

Therefore, the equivalent fraction is $\mathbf{\frac{3}{4}}$.

Solution:

Two fractions $\frac{a}{b}$ and $\frac{c}{d}$ are equivalent if their cross products are equal, i.e., if $a \times d = b \times c$. Another method is to simplify both fractions to their lowest terms; if the simplified forms are identical, the fractions are equivalent.

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(a) $\frac{5}{9}$ and $\frac{30}{54}$

Method 1: Cross-Multiplication

We check if $5 \times 54 = 9 \times 30$.

$5 \times 54 = 270$

$9 \times 30 = 270$

Since the cross products are equal ($270 = 270$), the fractions $\mathbf{\frac{5}{9}}$ and $\mathbf{\frac{30}{54}}$ are equivalent.

Method 2: Simplification

The fraction $\frac{5}{9}$ is already in its simplest form.

Simplify $\frac{30}{54}$. The greatest common divisor (GCD) of 30 and 54 is 6.

$\frac{30 \div 6}{54 \div 6} = \frac{5}{9}$

Since both fractions simplify to $\frac{5}{9}$, they are equivalent.

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(b) $\frac{3}{10}$ and $\frac{12}{50}$

Method 1: Cross-Multiplication

We check if $3 \times 50 = 10 \times 12$.

$3 \times 50 = 150$

$10 \times 12 = 120$

Since the cross products are not equal ($150 \neq 120$), the fractions $\mathbf{\frac{3}{10}}$ and $\mathbf{\frac{12}{50}}$ are not equivalent.

Method 2: Simplification

The fraction $\frac{3}{10}$ is already in its simplest form.

Simplify $\frac{12}{50}$. The GCD of 12 and 50 is 2.

$\frac{12 \div 2}{50 \div 2} = \frac{6}{25}$

Since $\frac{3}{10}$ is not the same as $\frac{6}{25}$, the fractions are not equivalent.

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(c) $\frac{7}{13}$ and $\frac{5}{11}$

Method 1: Cross-Multiplication

We check if $7 \times 11 = 13 \times 5$.

$7 \times 11 = 77$

$13 \times 5 = 65$

Since the cross products are not equal ($77 \neq 65$), the fractions $\mathbf{\frac{7}{13}}$ and $\mathbf{\frac{5}{11}}$ are not equivalent.

Method 2: Simplification

The fraction $\frac{7}{13}$ is in its simplest form (7 and 13 are prime numbers).

The fraction $\frac{5}{11}$ is in its simplest form (5 and 11 are prime numbers).

Since their simplest forms are different, the fractions are not equivalent.

Solution:

To reduce a fraction to its simplest form, we divide both the numerator and the denominator by their Greatest Common Divisor (GCD).

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(a) $\frac{48}{60}$

Find the GCD of 48 and 60.

Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

The GCD(48, 60) is 12.

Divide both numerator and denominator by 12:

$\frac{48 \div 12}{60 \div 12} = \mathbf{\frac{4}{5}}$

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(b) $\frac{150}{60}$

Find the GCD of 150 and 60.

We can first divide by 10: $\frac{150}{60} = \frac{15}{6}$

Now find the GCD of 15 and 6, which is 3.

Divide 15 and 6 by 3: $\frac{15 \div 3}{6 \div 3} = \frac{5}{2}$

(Alternatively, the GCD of 150 and 60 is 30.)

$\frac{150 \div 30}{60 \div 30} = \mathbf{\frac{5}{2}}$

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(c) $\frac{84}{98}$

Find the GCD of 84 and 98.

We can divide both by 2: $\frac{84 \div 2}{98 \div 2} = \frac{42}{49}$

Now find the GCD of 42 and 49, which is 7.

Divide 42 and 49 by 7: $\frac{42 \div 7}{49 \div 7} = \frac{6}{7}$

(Alternatively, the GCD of 84 and 98 is 14.)

$\frac{84 \div 14}{98 \div 14} = \mathbf{\frac{6}{7}}$

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(d) $\frac{12}{52}$

Find the GCD of 12 and 52.

Factors of 12: 1, 2, 3, 4, 6, 12

Factors of 52: 1, 2, 4, 13, 26, 52

The GCD(12, 52) is 4.

Divide both numerator and denominator by 4:

$\frac{12 \div 4}{52 \div 4} = \mathbf{\frac{3}{13}}$

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(e) $\frac{7}{28}$

Find the GCD of 7 and 28.

Since 7 divides 28 ($28 = 7 \times 4$), the GCD is 7.

Divide both numerator and denominator by 7:

$\frac{7 \div 7}{28 \div 7} = \mathbf{\frac{1}{4}}$

Solution:

We need to calculate the fraction of pencils used by each person and then compare these fractions.

The fraction of pencils used is calculated as: $\frac{\text{Number of pencils used}}{\text{Total number of pencils initially}}$.

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Ramesh:

Total pencils initially = 20.

Number of pencils used = 10.

Fraction used by Ramesh = $\frac{10}{20}$.

Simplifying the fraction:

$\frac{10 \div 10}{20 \div 10} = \mathbf{\frac{1}{2}}$.

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Sheelu:

Total pencils initially = 50.

Number of pencils used = 25.

Fraction used by Sheelu = $\frac{25}{50}$.

Simplifying the fraction:

$\frac{25 \div 25}{50 \div 25} = \mathbf{\frac{1}{2}}$.

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Jamaal:

Total pencils initially = 80.

Number of pencils used = 40.

Fraction used by Jamaal = $\frac{40}{80}$.

Simplifying the fraction:

$\frac{40 \div 40}{80 \div 40} = \mathbf{\frac{1}{2}}$.

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Comparison:

The fraction of pencils used by Ramesh is $\frac{1}{2}$.

The fraction of pencils used by Sheelu is $\frac{1}{2}$.

The fraction of pencils used by Jamaal is $\frac{1}{2}$.

Since all the simplified fractions are equal ($\frac{1}{2} = \frac{1}{2} = \frac{1}{2}$), yes, each has used up an equal fraction of their pencils.

Solution:

First, we simplify each fraction in the first column (i to v) to its simplest form. Then we match it with the fractions in the second column (a to e). Finally, we write two more equivalent fractions for each.

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(i) $\frac{250}{400}$

Divide numerator and denominator by their GCD. We can start by dividing by 10:

$\frac{250 \div 10}{400 \div 10} = \frac{25}{40}$

Now, divide by their GCD, which is 5:

$\frac{25 \div 5}{40 \div 5} = \mathbf{\frac{5}{8}}$

This matches with (d) $\frac{5}{8}$.

Two more equivalent fractions for $\frac{5}{8}$:

$\frac{5 \times 2}{8 \times 2} = \frac{10}{16}$

$\frac{5 \times 3}{8 \times 3} = \frac{15}{24}$

So, (i) $\leftrightarrow$ (d). Two more: $\frac{10}{16}$, $\frac{15}{24}$.

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(ii) $\frac{180}{200}$

Divide by 10:

$\frac{180 \div 10}{200 \div 10} = \frac{18}{20}$

Now, divide by their GCD, which is 2:

$\frac{18 \div 2}{20 \div 2} = \mathbf{\frac{9}{10}}$

This matches with (e) $\frac{9}{10}$.

Two more equivalent fractions for $\frac{9}{10}$:

$\frac{9 \times 2}{10 \times 2} = \frac{18}{20}$

$\frac{9 \times 3}{10 \times 3} = \frac{27}{30}$

So, (ii) $\leftrightarrow$ (e). Two more: $\frac{18}{20}$, $\frac{27}{30}$.

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(iii) $\frac{660}{990}$

Divide by 10:

$\frac{660 \div 10}{990 \div 10} = \frac{66}{99}$

Now, divide by their GCD, which is 33:

$\frac{66 \div 33}{99 \div 33} = \mathbf{\frac{2}{3}}$

This matches with (a) $\frac{2}{3}$.

Two more equivalent fractions for $\frac{2}{3}$:

$\frac{2 \times 2}{3 \times 2} = \frac{4}{6}$

$\frac{2 \times 4}{3 \times 4} = \frac{8}{12}$

So, (iii) $\leftrightarrow$ (a). Two more: $\frac{4}{6}$, $\frac{8}{12}$.

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(iv) $\frac{180}{360}$

Divide by 10:

$\frac{180 \div 10}{360 \div 10} = \frac{18}{36}$

Now, divide by their GCD, which is 18:

$\frac{18 \div 18}{36 \div 18} = \mathbf{\frac{1}{2}}$

This matches with (c) $\frac{1}{2}$.

Two more equivalent fractions for $\frac{1}{2}$:

$\frac{1 \times 3}{2 \times 3} = \frac{3}{6}$

$\frac{1 \times 4}{2 \times 4} = \frac{4}{8}$

So, (iv) $\leftrightarrow$ (c). Two more: $\frac{3}{6}$, $\frac{4}{8}$.

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(v) $\frac{220}{550}$

Divide by 10:

$\frac{220 \div 10}{550 \div 10} = \frac{22}{55}$

Now, divide by their GCD, which is 11:

$\frac{22 \div 11}{55 \div 11} = \mathbf{\frac{2}{5}}$

This matches with (b) $\frac{2}{5}$.

Two more equivalent fractions for $\frac{2}{5}$:

$\frac{2 \times 2}{5 \times 2} = \frac{4}{10}$

$\frac{2 \times 3}{5 \times 3} = \frac{6}{15}$

So, (v) $\leftrightarrow$ (b). Two more: $\frac{4}{10}$, $\frac{6}{15}$.

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Summary of Matches:

(i) $\leftrightarrow$ (d)

(ii) $\leftrightarrow$ (e)

(iii) $\leftrightarrow$ (a)

(iv) $\leftrightarrow$ (c)

(v) $\leftrightarrow$ (b)

Solution:

To compare the fractions $\frac{4}{5}$ and $\frac{5}{6}$, we need to convert them into like fractions (fractions with the same denominator).

Method 1: Finding a Common Denominator

1. Find the Least Common Multiple (LCM) of the denominators 5 and 6.

Since 5 and 6 have no common factors other than 1, their LCM is their product.

LCM(5, 6) = $5 \times 6 = 30$.

2. Convert each fraction to an equivalent fraction with a denominator of 30.

For $\frac{4}{5}$: To get 30 in the denominator, we multiply 5 by 6 ($30 \div 5 = 6$). So, we multiply the numerator by 6 as well.

$\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}$.

For $\frac{5}{6}$: To get 30 in the denominator, we multiply 6 by 5 ($30 \div 6 = 5$). So, we multiply the numerator by 5 as well.

$\frac{5}{6} = \frac{5 \times 5}{6 \times 5} = \frac{25}{30}$.

3. Now, compare the like fractions $\frac{24}{30}$ and $\frac{25}{30}$.

Since the denominators are the same, we compare the numerators.

$24 < 25$.

Therefore, $\frac{24}{30} < \frac{25}{30}$.

4. This means the original fractions have the same relationship:

$\mathbf{\frac{4}{5} < \frac{5}{6}}$.

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Method 2: Cross-Multiplication

To compare $\frac{a}{b}$ and $\frac{c}{d}$, we can compare the products $a \times d$ and $b \times c$.

For $\frac{4}{5}$ and $\frac{5}{6}$:

Calculate $4 \times 6 = 24$.

Calculate $5 \times 5 = 25$.

Since $24 < 25$, the first fraction is less than the second fraction.

Therefore, $\mathbf{\frac{4}{5} < \frac{5}{6}}$.

Solution:

We need to compare the fractions $\frac{5}{6}$ and $\frac{13}{15}$. We can do this by converting them to like fractions or by using the cross-multiplication method.

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Method 1: Finding a Common Denominator (LCM Method)

1. Find the Least Common Multiple (LCM) of the denominators 6 and 15.

Prime factorization of 6 = $2 \times 3$.

Prime factorization of 15 = $3 \times 5$.

LCM(6, 15) = $2 \times 3 \times 5 = 30$.

2. Convert each fraction to an equivalent fraction with the denominator 30.

For $\frac{5}{6}$: Since $30 \div 6 = 5$, we multiply the numerator and denominator by 5.

$\frac{5}{6} = \frac{5 \times 5}{6 \times 5} = \frac{25}{30}$.

For $\frac{13}{15}$: Since $30 \div 15 = 2$, we multiply the numerator and denominator by 2.

$\frac{13}{15} = \frac{13 \times 2}{15 \times 2} = \frac{26}{30}$.

3. Now, compare the like fractions $\frac{25}{30}$ and $\frac{26}{30}$.

Since the denominators are the same (30), we compare the numerators.

We see that $25 < 26$.

Therefore, $\frac{25}{30} < \frac{26}{30}$.

4. This implies that the original fractions have the same relationship:

$\mathbf{\frac{5}{6} < \frac{13}{15}}$.

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Method 2: Cross-Multiplication

To compare $\frac{a}{b}$ and $\frac{c}{d}$, we compare the products $a \times d$ and $b \times c$.

For $\frac{5}{6}$ and $\frac{13}{15}$:

Calculate the first cross product: $5 \times 15 = 75$.

Calculate the second cross product: $6 \times 13 = 78$.

Compare the products: $75$ and $78$.

Since $75 < 78$, the fraction corresponding to the numerator in the first product ($\frac{5}{6}$) is less than the fraction corresponding to the numerator in the second product ($\frac{13}{15}$).

Therefore, $\mathbf{\frac{5}{6} < \frac{13}{15}}$.

Solution:

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(a) Fractions from the first set of figures:

- Figure 1: Total parts = 8, Shaded parts = 3. Fraction = $\frac{3}{8}$.

- Figure 2: Total parts = 8, Shaded parts = 6. Fraction = $\frac{6}{8}$.

- Figure 3: Total parts = 8, Shaded parts = 4. Fraction = $\frac{4}{8}$.

- Figure 4: Total parts = 8, Shaded parts = 1. Fraction = $\frac{1}{8}$.

The fractions are $\frac{3}{8}$, $\frac{6}{8}$, $\frac{4}{8}$, $\frac{1}{8}$.

These are like fractions (same denominator). We compare their numerators: 1, 3, 4, 6.

Ascending Order: Comparing numerators $1 < 3 < 4 < 6$.

Therefore, $\mathbf{\frac{1}{8} < \frac{3}{8} < \frac{4}{8} < \frac{6}{8}}$.

Descending Order: Comparing numerators $6 > 4 > 3 > 1$.

Therefore, $\mathbf{\frac{6}{8} > \frac{4}{8} > \frac{3}{8} > \frac{1}{8}}$.

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(b) Fractions from the second set of figures:

- Figure 1: Total parts = 9, Shaded parts = 8. Fraction = $\frac{8}{9}$.

- Figure 2: Total parts = 9, Shaded parts = 4. Fraction = $\frac{4}{9}$.

- Figure 3: Total parts = 9, Shaded parts = 3. Fraction = $\frac{3}{9}$.

- Figure 4: Total parts = 9, Shaded parts = 6. Fraction = $\frac{6}{9}$.

The fractions are $\frac{8}{9}$, $\frac{4}{9}$, $\frac{3}{9}$, $\frac{6}{9}$.

These are like fractions (same denominator). We compare their numerators: 3, 4, 6, 8.

Ascending Order: Comparing numerators $3 < 4 < 6 < 8$.

Therefore, $\mathbf{\frac{3}{9} < \frac{4}{9} < \frac{6}{9} < \frac{8}{9}}$.

Descending Order: Comparing numerators $8 > 6 > 4 > 3$.

Therefore, $\mathbf{\frac{8}{9} > \frac{6}{9} > \frac{4}{9} > \frac{3}{9}}$.

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(c) Number line representation and comparison:

We need to show $\frac{2}{6}$, $\frac{4}{6}$, $\frac{8}{6}$, and $\frac{6}{6}$ on the number line. The denominator is 6, so we divide the intervals between integers into 6 equal parts.

Note that $\frac{6}{6} = 1$ and $\frac{8}{6} = 1\frac{2}{6}$.

$\ \ \ \ \ \ \ \ \frac{2}{6}\ \ \ \ \ \frac{4}{6}\ \ \ \ \ \frac{6}{6}\ \ \ \ \ \frac{8}{6}$

Now, put appropriate signs between the given fractions:

- $\frac{5}{6}$ vs $\frac{2}{6}$: Since $5 > 2$, $\mathbf{\frac{5}{6} > \frac{2}{6}}$.

- $\frac{3}{6}$ vs $0$: Since $0 = \frac{0}{6}$ and $3 > 0$, $\mathbf{\frac{3}{6} > 0}$.

- $\frac{1}{6}$ vs $\frac{6}{6}$: Since $1 < 6$, $\mathbf{\frac{1}{6} < \frac{6}{6}}$.

- $\frac{8}{6}$ vs $\frac{5}{6}$: Since $8 > 5$, $\mathbf{\frac{8}{6} > \frac{5}{6}}$.

Completed comparisons:

Solution:

We will compare each pair of fractions using the rules for comparing fractions.

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(a) $\frac{3}{6}$ vs $\frac{5}{6}$

These are like fractions (same denominator = 6).

When denominators are the same, the fraction with the larger numerator is the larger fraction.

Compare the numerators: $3 < 5$.

Therefore, $\frac{3}{6}$ < $\frac{5}{6}$.

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(b) $\frac{1}{7}$ vs $\frac{1}{4}$

These fractions have the same numerator (numerator = 1).

When numerators are the same, the fraction with the smaller denominator is the larger fraction (because the whole is divided into fewer, larger pieces).

Compare the denominators: $7 > 4$.

The fraction with the smaller denominator (4) is larger.

Therefore, $\frac{1}{7}$ < $\frac{1}{4}$.

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(c) $\frac{4}{5}$ vs $\frac{5}{5}$

These are like fractions (same denominator = 5).

Compare the numerators: $4 < 5$.

Therefore, $\frac{4}{5}$ < $\frac{5}{5}$.

(Alternatively, note that $\frac{5}{5} = 1$. Since the numerator 4 is less than the denominator 5, $\frac{4}{5}$ is a proper fraction and is less than 1.)

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(d) $\frac{3}{5}$ vs $\frac{3}{7}$

These fractions have the same numerator (numerator = 3).

When numerators are the same, the fraction with the smaller denominator is the larger fraction.

Compare the denominators: $5 < 7$.

The fraction with the smaller denominator (5) is larger.

Therefore, $\frac{3}{5}$ > $\frac{3}{7}$.

Solution:

Here are five pairs of fractions with appropriate comparison signs:

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Pair 1: $\frac{2}{9}$ vs $\frac{7}{9}$

These are like fractions (denominator = 9). Comparing numerators: $2 < 7$.

Result: $\mathbf{\frac{2}{9} < \frac{7}{9}}$

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Pair 2: $\frac{4}{7}$ vs $\frac{4}{5}$

These fractions have the same numerator (numerator = 4). Comparing denominators: $7 > 5$. The fraction with the smaller denominator is larger.

Result: $\mathbf{\frac{4}{7} < \frac{4}{5}}$

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Pair 3: $\frac{2}{3}$ vs $\frac{3}{5}$

These are unlike fractions. We can use the LCM method. LCM(3, 5) = 15.

$\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$

$\frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15}$

Comparing $\frac{10}{15}$ and $\frac{9}{15}$: Since $10 > 9$, $\frac{10}{15} > \frac{9}{15}$.

Result: $\mathbf{\frac{2}{3} > \frac{3}{5}}$

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Pair 4: $\frac{3}{12}$ vs $\frac{2}{8}$

We can simplify both fractions.

$\frac{3}{12} = \frac{3 \div 3}{12 \div 3} = \frac{1}{4}$

$\frac{2}{8} = \frac{2 \div 2}{8 \div 2} = \frac{1}{4}$

Since both simplify to $\frac{1}{4}$, they are equal.

Result: $\mathbf{\frac{3}{12} = \frac{2}{8}}$

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Pair 5: $\frac{5}{8}$ vs $\frac{7}{11}$

These are unlike fractions. We can use the cross-multiplication method.

Compare $5 \times 11$ and $8 \times 7$.

$5 \times 11 = 55$

$8 \times 7 = 56$

Since $55 < 56$, the first fraction is smaller than the second fraction.

Result: $\mathbf{\frac{5}{8} < \frac{7}{11}}$

Solution:

By observing the positions of the fractions on the provided number lines (fraction strips):

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(a) $\frac{1}{6}$ vs $\frac{1}{3}$

Locating $\frac{1}{6}$ on the sixths strip and $\frac{1}{3}$ on the thirds strip, we see that $\frac{1}{6}$ is to the left of $\frac{1}{3}$.

Alternatively, comparing fractions with the same numerator (1), the one with the larger denominator (6) is smaller than the one with the smaller denominator (3).

Therefore, $\mathbf{\frac{1}{6} < \frac{1}{3}}$.

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(b) $\frac{3}{4}$ vs $\frac{2}{6}$

Locating $\frac{3}{4}$ on the fourths strip and $\frac{2}{6}$ on the sixths strip, we see that $\frac{3}{4}$ is to the right of $\frac{2}{6}$.

Alternatively, simplify $\frac{2}{6} = \frac{1}{3}$. Compare $\frac{3}{4}$ and $\frac{1}{3}$. LCM(4, 3) = 12. $\frac{3}{4} = \frac{9}{12}$ and $\frac{1}{3} = \frac{4}{12}$. Since $9 > 4$, $\frac{9}{12} > \frac{4}{12}$.

Therefore, $\mathbf{\frac{3}{4} > \frac{2}{6}}$.

---

(c) $\frac{2}{3}$ vs $\frac{2}{4}$

Locating $\frac{2}{3}$ on the thirds strip and $\frac{2}{4}$ on the fourths strip, we see that $\frac{2}{3}$ is to the right of $\frac{2}{4}$.

Alternatively, comparing fractions with the same numerator (2), the one with the smaller denominator (3) is larger than the one with the larger denominator (4).

Therefore, $\mathbf{\frac{2}{3} > \frac{2}{4}}$.

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(d) $\frac{6}{6}$ vs $\frac{3}{3}$

Locating $\frac{6}{6}$ on the sixths strip and $\frac{3}{3}$ on the thirds strip, we see they both align with the mark for 1.

Alternatively, $\frac{6}{6} = 1$ and $\frac{3}{3} = 1$.

Therefore, $\mathbf{\frac{6}{6} = \frac{3}{3}}$.

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(e) $\frac{5}{6}$ vs $\frac{5}{5}$

Locating $\frac{5}{6}$ on the sixths strip and $\frac{5}{5}$ on the fifths strip, we see that $\frac{5}{6}$ is to the left of $\frac{5}{5}$ (which is 1).

Alternatively, comparing fractions with the same numerator (5), the one with the larger denominator (6) is smaller than the one with the smaller denominator (5).

Therefore, $\mathbf{\frac{5}{6} < \frac{5}{5}}$.

---

Five More Problems:

1. Compare $\frac{1}{2}$ and $\frac{3}{5}$.

From strips: $\frac{1}{2}$ is left of $\frac{3}{5}$. $\frac{1}{2} = \frac{5}{10}$, $\frac{3}{5} = \frac{6}{10}$. $5 < 6$. So, $\mathbf{\frac{1}{2} < \frac{3}{5}}$.

2. Compare $\frac{2}{4}$ and $\frac{3}{6}$.

From strips: They align. Simplify: $\frac{2}{4} = \frac{1}{2}$, $\frac{3}{6} = \frac{1}{2}$. So, $\mathbf{\frac{2}{4} = \frac{3}{6}}$.

3. Compare $\frac{4}{5}$ and $\frac{5}{6}$.

From strips: $\frac{4}{5}$ is left of $\frac{5}{6}$. Cross-multiply: $4 \times 6 = 24$, $5 \times 5 = 25$. $24 < 25$. So, $\mathbf{\frac{4}{5} < \frac{5}{6}}$.

4. Compare $\frac{0}{2}$ and $\frac{0}{6}$.

From strips: Both align with 0. $\frac{0}{2} = 0$, $\frac{0}{6} = 0$. So, $\mathbf{\frac{0}{2} = \frac{0}{6}}$.

5. Compare $\frac{3}{3}$ and $\frac{4}{5}$.

From strips: $\frac{3}{3}=1$ is right of $\frac{4}{5}$. $\frac{3}{3}=1$. $\frac{4}{5}$ is a proper fraction, so less than 1. So, $\mathbf{\frac{3}{3} > \frac{4}{5}}$.

Solution:

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(a) $\frac{1}{2}$ vs $\frac{1}{5}$

Same numerator (1). Compare denominators: $2 < 5$. The fraction with the smaller denominator is larger.

$\mathbf{\frac{1}{2} > \frac{1}{5}}$

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(b) $\frac{2}{4}$ vs $\frac{3}{6}$

Simplify both fractions: $\frac{2}{4} = \frac{1}{2}$ and $\frac{3}{6} = \frac{1}{2}$.

$\mathbf{\frac{2}{4} = \frac{3}{6}}$

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(c) $\frac{3}{5}$ vs $\frac{2}{3}$

Cross-multiplication: $3 \times 3 = 9$ and $5 \times 2 = 10$. Since $9 < 10$.

$\mathbf{\frac{3}{5} < \frac{2}{3}}$

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(d) $\frac{3}{4}$ vs $\frac{2}{8}$

Simplify $\frac{2}{8} = \frac{1}{4}$. Compare $\frac{3}{4}$ and $\frac{1}{4}$. Like fractions. Compare numerators: $3 > 1$.

$\mathbf{\frac{3}{4} > \frac{2}{8}}$

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(e) $\frac{3}{5}$ vs $\frac{6}{5}$

Like fractions (denominator = 5). Compare numerators: $3 < 6$.

$\mathbf{\frac{3}{5} < \frac{6}{5}}$

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(f) $\frac{7}{9}$ vs $\frac{3}{9}$

Like fractions (denominator = 9). Compare numerators: $7 > 3$.

$\mathbf{\frac{7}{9} > \frac{3}{9}}$

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(g) $\frac{1}{4}$ vs $\frac{2}{8}$

Simplify $\frac{2}{8} = \frac{1}{4}$.

$\mathbf{\frac{1}{4} = \frac{2}{8}}$

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(h) $\frac{6}{10}$ vs $\frac{4}{5}$

Simplify $\frac{6}{10} = \frac{3}{5}$. Compare $\frac{3}{5}$ and $\frac{4}{5}$. Like fractions. Compare numerators: $3 < 4$.

$\mathbf{\frac{6}{10} < \frac{4}{5}}$

---

(i) $\frac{3}{4}$ vs $\frac{7}{8}$

Convert $\frac{3}{4}$ to an equivalent fraction with denominator 8: $\frac{3}{4} = \frac{3 \times 2}{4 \times 2} = \frac{6}{8}$. Compare $\frac{6}{8}$ and $\frac{7}{8}$. Like fractions. Compare numerators: $6 < 7$.

$\mathbf{\frac{3}{4} < \frac{7}{8}}$

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(j) $\frac{6}{10}$ vs $\frac{3}{5}$

Simplify $\frac{6}{10} = \frac{3}{5}$.

$\mathbf{\frac{6}{10} = \frac{3}{5}}$

---

(k) $\frac{5}{7}$ vs $\frac{15}{21}$

Simplify $\frac{15}{21} = \frac{15 \div 3}{21 \div 3} = \frac{5}{7}$.

$\mathbf{\frac{5}{7} = \frac{15}{21}}$

Solution:

To group the fractions, we need to reduce each one to its simplest form by dividing the numerator and denominator by their Greatest Common Divisor (GCD).

---

Simplification of each fraction:

(a) $\frac{2}{12} = \frac{2 \div 2}{12 \div 2} = \mathbf{\frac{1}{6}}$

(b) $\frac{3}{15} = \frac{3 \div 3}{15 \div 3} = \mathbf{\frac{1}{5}}$

(c) $\frac{8}{50} = \frac{8 \div 2}{50 \div 2} = \mathbf{\frac{4}{25}}$

(d) $\frac{16}{100} = \frac{16 \div 4}{100 \div 4} = \mathbf{\frac{4}{25}}$

(e) $\frac{10}{60} = \frac{10 \div 10}{60 \div 10} = \mathbf{\frac{1}{6}}$

(f) $\frac{15}{75} = \frac{15 \div 15}{75 \div 15} = \mathbf{\frac{1}{5}}$

(g) $\frac{12}{60} = \frac{12 \div 12}{60 \div 12} = \mathbf{\frac{1}{5}}$

(h) $\frac{16}{96} = \frac{16 \div 16}{96 \div 16} = \mathbf{\frac{1}{6}}$

(i) $\frac{12}{75} = \frac{12 \div 3}{75 \div 3} = \mathbf{\frac{4}{25}}$

(j) $\frac{12}{72} = \frac{12 \div 12}{72 \div 12} = \mathbf{\frac{1}{6}}$

(k) $\frac{3}{18} = \frac{3 \div 3}{18 \div 3} = \mathbf{\frac{1}{6}}$

(l) $\frac{4}{25}$ is already in its simplest form: $\mathbf{\frac{4}{25}}$

---

Grouping the equivalent fractions:

We group the original fractions based on their common simplest form.

Group 1 (Simplest form $\frac{1}{6}$):

(a) $\frac{2}{12}$, (e) $\frac{10}{60}$, (h) $\frac{16}{96}$, (j) $\frac{12}{72}$, (k) $\frac{3}{18}$

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Group 2 (Simplest form $\frac{1}{5}$):

(b) $\frac{3}{15}$, (f) $\frac{15}{75}$, (g) $\frac{12}{60}$

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Group 3 (Simplest form $\frac{4}{25}$):

(c) $\frac{8}{50}$, (d) $\frac{16}{100}$, (i) $\frac{12}{75}$, (l) $\frac{4}{25}$

Solution:

To check if two fractions are equal, we can use either the cross-multiplication method or simplify one or both fractions to their simplest forms.

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(a) Is $\frac{5}{9}$ equal to $\frac{4}{5}$ ?

Method: Cross-Multiplication

We check if the cross products are equal: $5 \times 5$ and $9 \times 4$.

$5 \times 5 = 25$.

$9 \times 4 = 36$.

Since the cross products are not equal ($25 \neq 36$), the fractions are not equal.

Answer: No, $\frac{5}{9}$ is not equal to $\frac{4}{5}$.

---

(b) Is $\frac{9}{16}$ equal to $\frac{5}{9}$ ?

Method: Cross-Multiplication

We check if the cross products are equal: $9 \times 9$ and $16 \times 5$.

$9 \times 9 = 81$.

$16 \times 5 = 80$.

Since the cross products are not equal ($81 \neq 80$), the fractions are not equal.

Answer: No, $\frac{9}{16}$ is not equal to $\frac{5}{9}$.

---

(c) Is $\frac{4}{5}$ equal to $\frac{16}{20}$ ?

Method: Simplification

The fraction $\frac{4}{5}$ is already in its simplest form.

Let's simplify $\frac{16}{20}$. The Greatest Common Divisor (GCD) of 16 and 20 is 4.

$\frac{16 \div 4}{20 \div 4} = \frac{4}{5}$.

Since the simplified form of $\frac{16}{20}$ is $\frac{4}{5}$, which is the same as the first fraction, the fractions are equal.

Answer: Yes, $\frac{4}{5}$ is equal to $\frac{16}{20}$.

---

(d) Is $\frac{1}{15}$ equal to $\frac{4}{30}$ ?

Method: Simplification

The fraction $\frac{1}{15}$ is already in its simplest form.

Let's simplify $\frac{4}{30}$. The GCD of 4 and 30 is 2.

$\frac{4 \div 2}{30 \div 2} = \frac{2}{15}$.

Since the simplest form of $\frac{4}{30}$ is $\frac{2}{15}$, which is not equal to $\frac{1}{15}$, the fractions are not equal.

Answer: No, $\frac{1}{15}$ is not equal to $\frac{4}{30}$.

Solution:

To determine who read less, we need to compare the fraction of the book each person read.

The total number of pages in the book = 100.

---

Fraction read by Ila:

Ila read 25 pages out of 100 pages.

Fraction = $\frac{\text{Pages read by Ila}}{\text{Total pages}} = \frac{25}{100}$.

Let's simplify this fraction to its lowest terms. The Greatest Common Divisor (GCD) of 25 and 100 is 25.

Simplified fraction for Ila = $\frac{25 \div 25}{100 \div 25} = \mathbf{\frac{1}{4}}$.

---

Fraction read by Lalita:

Lalita read $\mathbf{\frac{2}{5}}$ of the book.

---

Comparison:

Now, we need to compare the fractions $\frac{1}{4}$ (Ila) and $\frac{2}{5}$ (Lalita).

We can compare them by finding a common denominator or by cross-multiplication.

Method 1: Common Denominator (LCM)

The denominators are 4 and 5. The Least Common Multiple (LCM) of 4 and 5 is $4 \times 5 = 20$.

Convert $\frac{1}{4}$ to an equivalent fraction with denominator 20:

$\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$.

Convert $\frac{2}{5}$ to an equivalent fraction with denominator 20:

$\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}$.

Now compare the numerators: $5 < 8$.

So, $\frac{5}{20} < \frac{8}{20}$.

This means $\frac{1}{4} < \frac{2}{5}$.

Method 2: Cross-Multiplication

Compare $\frac{1}{4}$ and $\frac{2}{5}$.

Calculate $1 \times 5 = 5$.

Calculate $4 \times 2 = 8$.

Since $5 < 8$, the first fraction is smaller than the second fraction.

So, $\frac{1}{4} < \frac{2}{5}$.

---

Conclusion:

Since the fraction read by Ila ($\frac{1}{4}$) is less than the fraction read by Lalita ($\frac{2}{5}$), Ila read less.

Solution:

We need to compare the time duration for which Rafiq and Rohit exercised.

Time exercised by Rafiq = $\frac{3}{6}$ of an hour.

Time exercised by Rohit = $\frac{3}{4}$ of an hour.

To find out who exercised longer, we need to compare the fractions $\frac{3}{6}$ and $\frac{3}{4}$.

---

Method 1: Using the rule for fractions with the same numerator.

The fractions are $\frac{3}{6}$ and $\frac{3}{4}$. They have the same numerator, which is 3.

When comparing fractions with the same numerator, the fraction with the smaller denominator is the larger fraction.

Compare the denominators: $6$ and $4$.

Since $4 < 6$, the fraction $\frac{3}{4}$ is greater than the fraction $\frac{3}{6}$.

So, $\frac{3}{4} > \frac{3}{6}$.

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Method 2: Simplifying and Finding a Common Denominator.

First, simplify $\frac{3}{6}$. The GCD of 3 and 6 is 3.

$\frac{3}{6} = \frac{3 \div 3}{6 \div 3} = \frac{1}{2}$.

Now compare $\frac{1}{2}$ and $\frac{3}{4}$.

Find the LCM of the denominators 2 and 4. LCM(2, 4) = 4.

Convert $\frac{1}{2}$ to an equivalent fraction with denominator 4:

$\frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{4}$.

Compare the like fractions $\frac{2}{4}$ and $\frac{3}{4}$.

Since the denominators are the same, compare the numerators: $2 < 3$.

Therefore, $\frac{2}{4} < \frac{3}{4}$, which means $\frac{3}{6} < \frac{3}{4}$.

---

Conclusion:

Since $\frac{3}{4}$ is greater than $\frac{3}{6}$, Rohit exercised for a longer time than Rafiq.

Therefore, Rohit exercised for a longer time.

Solution:

We need to calculate the fraction of students who passed with 60% or more marks for each class and then compare these fractions.

Fraction = $\frac{\text{Number of students passed with 60% or more}}{\text{Total number of students in the class}}$

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Class A:

Total number of students = 25.

Number of students passed with 60% or more = 20.

Fraction for Class A = $\frac{20}{25}$.

Simplify the fraction. The Greatest Common Divisor (GCD) of 20 and 25 is 5.

Fraction for Class A = $\frac{20 \div 5}{25 \div 5} = \mathbf{\frac{4}{5}}$.

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Class B:

Total number of students = 30.

Number of students passed with 60% or more = 24.

Fraction for Class B = $\frac{24}{30}$.

Simplify the fraction. The GCD of 24 and 30 is 6.

Fraction for Class B = $\frac{24 \div 6}{30 \div 6} = \mathbf{\frac{4}{5}}$.

---

Comparison:

Fraction for Class A = $\frac{4}{5}$.

Fraction for Class B = $\frac{4}{5}$.

Comparing the two fractions, we find that $\frac{4}{5} = \frac{4}{5}$.

---

Conclusion:

Both classes had the same fraction of students getting 60% or more marks.

Solution:

We need to represent the operations shown in the figures using fractions.

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(a)

First figure: Represents 1 shaded part out of 5 total parts. Fraction = $\frac{1}{5}$.

Second figure: Represents 2 shaded parts out of 5 total parts. Fraction = $\frac{2}{5}$.

Third figure: Represents 3 shaded parts out of 5 total parts. Fraction = $\frac{3}{5}$.

The figures show that the shaded parts from the first two figures are combined in the third figure ($1+2=3$).

This represents an addition operation.

Appropriate representation: $\mathbf{\frac{1}{5} + \frac{2}{5} = \frac{3}{5}}$

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(b)

First figure: Represents 5 shaded parts out of 5 total parts (the whole). Fraction = $\frac{5}{5}$.

Second figure: Represents 3 shaded parts out of 5 total parts. Fraction = $\frac{3}{5}$.

Third figure: Represents 2 shaded parts out of 5 total parts. Fraction = $\frac{2}{5}$.

The figures show that 3 shaded parts are removed from the initial 5 shaded parts, leaving 2 shaded parts ($5-3=2$).

This represents a subtraction operation.

Appropriate representation: $\mathbf{\frac{5}{5} - \frac{3}{5} = \frac{2}{5}}$

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(c)

First figure: Represents 2 shaded parts out of 6 total parts. Fraction = $\frac{2}{6}$.

Second figure: Represents 3 shaded parts out of 6 total parts. Fraction = $\frac{3}{6}$.

Third figure: Represents 5 shaded parts out of 6 total parts. Fraction = $\frac{5}{6}$.

The figures show that the shaded parts from the first two figures are combined in the third figure ($2+3=5$).

This represents an addition operation.

Appropriate representation: $\mathbf{\frac{2}{6} + \frac{3}{6} = \frac{5}{6}}$

Solution:

We solve the addition and subtraction of fractions. For like fractions (fractions with the same denominator), we add or subtract the numerators and keep the denominator the same.

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(a) $\frac{1}{18} + \frac{1}{18}$

These are like fractions.

$\frac{1}{18} + \frac{1}{18} = \frac{1+1}{18} = \frac{2}{18}$

Simplify the result: $\frac{2 \div 2}{18 \div 2} = \mathbf{\frac{1}{9}}$

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(b) $\frac{8}{15} + \frac{3}{15}$

These are like fractions.

$\frac{8}{15} + \frac{3}{15} = \frac{8+3}{15} = \mathbf{\frac{11}{15}}$

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(c) $\frac{7}{7} - \frac{5}{7}$

These are like fractions.

$\frac{7}{7} - \frac{5}{7} = \frac{7-5}{7} = \mathbf{\frac{2}{7}}$

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(d) $\frac{1}{22} + \frac{21}{22}$

These are like fractions.

$\frac{1}{22} + \frac{21}{22} = \frac{1+21}{22} = \frac{22}{22} = \mathbf{1}$

---

(e) $\frac{12}{15} - \frac{7}{15}$

These are like fractions.

$\frac{12}{15} - \frac{7}{15} = \frac{12-7}{15} = \frac{5}{15}$

Simplify the result: $\frac{5 \div 5}{15 \div 5} = \mathbf{\frac{1}{3}}$

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(f) $\frac{5}{8} + \frac{3}{8}$

These are like fractions.

$\frac{5}{8} + \frac{3}{8} = \frac{5+3}{8} = \frac{8}{8} = \mathbf{1}$

---

(g) $1 - \frac{2}{3}$

First, write 1 as a fraction with denominator 3: $1 = \frac{3}{3}$.

Now subtract the like fractions:

$\frac{3}{3} - \frac{2}{3} = \frac{3-2}{3} = \mathbf{\frac{1}{3}}$

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(h) $\frac{1}{4} + \frac{0}{4}$

These are like fractions.

$\frac{1}{4} + \frac{0}{4} = \frac{1+0}{4} = \mathbf{\frac{1}{4}}$

---

(i) $3 - \frac{12}{5}$

First, write 3 as a fraction with denominator 5: $3 = \frac{3 \times 5}{5} = \frac{15}{5}$.

Now subtract the like fractions:

$\frac{15}{5} - \frac{12}{5} = \frac{15-12}{5} = \mathbf{\frac{3}{5}}$

Solution:

To find the total portion of the wall painted by Shubham and Madhavi together, we need to add the fractions of the wall space each of them painted.

Fraction of wall painted by Shubham = $\frac{2}{3}$.

Fraction of wall painted by Madhavi = $\frac{1}{3}$.

Total fraction painted together = Fraction painted by Shubham + Fraction painted by Madhavi

Total fraction = $\frac{2}{3} + \frac{1}{3}$.

Since these are like fractions (they have the same denominator, 3), we can add their numerators directly and keep the common denominator:

Total fraction = $\frac{2 + 1}{3} = \frac{3}{3}$.

Simplify the resulting fraction:

$\frac{3}{3} = 1$.

Therefore, Shubham and Madhavi painted the entire wall space (1 whole) together.

Solution:

To find the missing fraction, we can represent the unknown fraction with a variable or think about the relationship between the given fractions.

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(a) $\frac{7}{10} - \boxed{?} = \frac{3}{10}$

Let the missing fraction be $\frac{x}{10}$.

The equation is $\frac{7}{10} - \frac{x}{10} = \frac{3}{10}$.

Since the denominators are the same, we can write the equation for the numerators: $7 - x = 3$.

To find $x$, we can rearrange: $x = 7 - 3 = 4$.

So, the missing fraction is $\mathbf{\frac{4}{10}}$.

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(b) $\boxed{?} - \frac{3}{21} = \frac{5}{21}$

Let the missing fraction be $\frac{y}{21}$.

The equation is $\frac{y}{21} - \frac{3}{21} = \frac{5}{21}$.

Writing the equation for the numerators: $y - 3 = 5$.

To find $y$, we add 3 to both sides: $y = 5 + 3 = 8$.

So, the missing fraction is $\mathbf{\frac{8}{21}}$.

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(c) $\boxed{?} - \frac{3}{6} = \frac{3}{6}$

Let the missing fraction be $\frac{z}{6}$.

The equation is $\frac{z}{6} - \frac{3}{6} = \frac{3}{6}$.

Writing the equation for the numerators: $z - 3 = 3$.

To find $z$, we add 3 to both sides: $z = 3 + 3 = 6$.

So, the missing fraction is $\mathbf{\frac{6}{6}}$ (or 1).

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(d) $\boxed{?} + \frac{5}{27} = \frac{12}{27}$

Let the missing fraction be $\frac{w}{27}$.

The equation is $\frac{w}{27} + \frac{5}{27} = \frac{12}{27}$.

Writing the equation for the numerators: $w + 5 = 12$.

To find $w$, we subtract 5 from both sides: $w = 12 - 5 = 7$.

So, the missing fraction is $\mathbf{\frac{7}{27}}$.

Solution:

Let the total basket of oranges represent the whole, which is equal to 1.

Fraction of oranges given to Javed = $\frac{5}{7}$.

To find the fraction of oranges left in the basket, we need to subtract the fraction given to Javed from the total (the whole basket).

Fraction left = Total fraction - Fraction given to Javed

Fraction left = $1 - \frac{5}{7}$.

To perform the subtraction, we need to express 1 as a fraction with the same denominator as $\frac{5}{7}$. The denominator is 7.

So, $1 = \frac{7}{7}$.

Now, substitute this into the subtraction:

Fraction left = $\frac{7}{7} - \frac{5}{7}$.

Since these are like fractions, subtract the numerators and keep the denominator:

Fraction left = $\frac{7 - 5}{7} = \frac{2}{7}$.

Therefore, the fraction of oranges left in the basket was $\mathbf{\frac{2}{7}}$.

Solution:

We need to calculate the difference: $\frac{5}{6} - \frac{3}{4}$.

These are unlike fractions because they have different denominators (6 and 4).

To subtract them, we first need to find a common denominator. The most efficient common denominator is the Least Common Multiple (LCM) of the denominators 6 and 4.

Multiples of 6 are 6, 12, 18, ...

Multiples of 4 are 4, 8, 12, 16, ...

The LCM(6, 4) is 12.

---

Now, we convert each fraction to an equivalent fraction with the denominator 12:

For $\frac{5}{6}$: We need to multiply the denominator 6 by 2 to get 12 ($12 \div 6 = 2$). So, we multiply the numerator 5 by 2 as well.

$\frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}$.

For $\frac{3}{4}$: We need to multiply the denominator 4 by 3 to get 12 ($12 \div 4 = 3$). So, we multiply the numerator 3 by 3 as well.

$\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}$.

---

Now, perform the subtraction using the equivalent like fractions:

$\frac{5}{6} - \frac{3}{4} = \frac{10}{12} - \frac{9}{12}$.

Subtract the numerators and keep the common denominator:

$\frac{10 - 9}{12} = \frac{1}{12}$.

The fraction $\frac{1}{12}$ is already in its simplest form.

Therefore, subtracting $\frac{3}{4}$ from $\frac{5}{6}$ gives $\mathbf{\frac{1}{12}}$.

Solution:

We need to calculate the sum: $\frac{2}{5} + \frac{1}{3}$.

These fractions are unlike fractions because their denominators (5 and 3) are different.

To add them, we must first convert them into like fractions by finding a common denominator. The best common denominator to use is the Least Common Multiple (LCM) of the denominators 5 and 3.

Since 5 and 3 are prime numbers, their LCM is their product:

LCM(5, 3) = $5 \times 3 = 15$.

---

Now, convert each fraction to an equivalent fraction with a denominator of 15:

For $\frac{2}{5}$: To get a denominator of 15, we multiply the denominator 5 by 3 ($15 \div 5 = 3$). So, we also multiply the numerator 2 by 3.

$\frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15}$.

For $\frac{1}{3}$: To get a denominator of 15, we multiply the denominator 3 by 5 ($15 \div 3 = 5$). So, we also multiply the numerator 1 by 5.

$\frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}$.

---

Now, add the equivalent like fractions:

$\frac{2}{5} + \frac{1}{3} = \frac{6}{15} + \frac{5}{15}$.

Add the numerators and keep the common denominator:

$\frac{6 + 5}{15} = \frac{11}{15}$.

The fraction $\frac{11}{15}$ is already in its simplest form because 11 and 15 have no common factors other than 1.

Therefore, the sum of $\frac{2}{5}$ and $\frac{1}{3}$ is $\mathbf{\frac{11}{15}}$.

Solution:

We need to simplify the expression $\frac{3}{5} - \frac{7}{20}$.

The given fractions, $\frac{3}{5}$ and $\frac{7}{20}$, are unlike fractions as they have different denominators (5 and 20).

To perform the subtraction, we first need to convert them into like fractions by finding a common denominator. We will use the Least Common Multiple (LCM) of the denominators 5 and 20.

Find the LCM of 5 and 20:

Multiples of 5 are 5, 10, 15, 20, 25, ...

Multiples of 20 are 20, 40, ...

The LCM(5, 20) is 20.

---

Now, convert the fraction $\frac{3}{5}$ into an equivalent fraction with the denominator 20. The fraction $\frac{7}{20}$ already has the denominator 20.

For $\frac{3}{5}$: To get a denominator of 20, we multiply the denominator 5 by 4 ($20 \div 5 = 4$). Therefore, we must also multiply the numerator 3 by 4.

$\frac{3}{5} = \frac{3 \times 4}{5 \times 4} = \frac{12}{20}$.

---

Now substitute the equivalent fraction back into the expression:

$\frac{3}{5} - \frac{7}{20} = \frac{12}{20} - \frac{7}{20}$.

Now we have like fractions. Subtract the numerators and keep the common denominator:

$\frac{12 - 7}{20} = \frac{5}{20}$.

---

Finally, simplify the resulting fraction $\frac{5}{20}$ to its lowest terms.

Find the Greatest Common Divisor (GCD) of 5 and 20. The GCD is 5.

Divide both the numerator and the denominator by 5:

$\frac{5 \div 5}{20 \div 5} = \mathbf{\frac{1}{4}}$.

Therefore, the simplified result of $\frac{3}{5} - \frac{7}{20}$ is $\mathbf{\frac{1}{4}}$.

Solution:

We need to calculate the sum $2\frac{4}{5} + 3\frac{5}{6}$.

Method 1: Converting to Improper Fractions

1. Convert the mixed fractions into improper fractions.

$2\frac{4}{5} = \frac{(2 \times 5) + 4}{5} = \frac{10 + 4}{5} = \frac{14}{5}$.

$3\frac{5}{6} = \frac{(3 \times 6) + 5}{6} = \frac{18 + 5}{6} = \frac{23}{6}$.

2. Now add the improper fractions: $\frac{14}{5} + \frac{23}{6}$.

3. Find the Least Common Multiple (LCM) of the denominators 5 and 6. Since 5 and 6 are coprime (their only common factor is 1), their LCM is their product.

LCM(5, 6) = $5 \times 6 = 30$.

4. Convert each improper fraction to an equivalent fraction with the denominator 30.

$\frac{14}{5} = \frac{14 \times 6}{5 \times 6} = \frac{84}{30}$.

$\frac{23}{6} = \frac{23 \times 5}{6 \times 5} = \frac{115}{30}$.

5. Add the resulting like fractions:

$\frac{84}{30} + \frac{115}{30} = \frac{84 + 115}{30} = \frac{199}{30}$.

6. Convert the improper fraction sum $\frac{199}{30}$ back to a mixed fraction.

Divide 199 by 30:

$\begin{array}{r} 6\phantom{)} \\ 30{\overline{\smash{\big)}\,199\phantom{)}}} \\ \underline{-~\phantom{(}180} \\ 19\phantom{)} \end{array}$

The quotient is 6, and the remainder is 19.

So, $\frac{199}{30} = \mathbf{6\frac{19}{30}}$.

---

Alternate Method: Adding Whole and Fractional Parts Separately

1. Add the whole number parts: $2 + 3 = 5$.

2. Add the fractional parts: $\frac{4}{5} + \frac{5}{6}$.

3. Find the LCM of the denominators 5 and 6, which is 30.

4. Convert the fractional parts to equivalent fractions with denominator 30:

$\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30}$.

$\frac{5}{6} = \frac{5 \times 5}{6 \times 5} = \frac{25}{30}$.

5. Add the like fractions:

$\frac{24}{30} + \frac{25}{30} = \frac{24 + 25}{30} = \frac{49}{30}$.

6. Convert the improper fraction sum $\frac{49}{30}$ to a mixed fraction:

$49 \div 30 = 1$ with a remainder of $19$. So, $\frac{49}{30} = 1\frac{19}{30}$.

7. Combine the sum of the whole parts (5) and the mixed fraction from the sum of the fractional parts ($1\frac{19}{30}$):

$5 + 1\frac{19}{30} = (5 + 1) + \frac{19}{30} = \mathbf{6\frac{19}{30}}$.

---

Both methods yield the same result. The sum is $\mathbf{6\frac{19}{30}}$.

Solution:

We need to calculate the difference $4\frac{2}{5} - 2\frac{1}{5}$.

Method 1: Subtracting Whole and Fractional Parts Separately

Since the fractional parts ($\frac{2}{5}$ and $\frac{1}{5}$) have the same denominator and the first fraction's fractional part ($\frac{2}{5}$) is greater than the second fraction's fractional part ($\frac{1}{5}$), we can subtract the whole parts and the fractional parts separately.

1. Subtract the whole number parts: $4 - 2 = 2$.

2. Subtract the fractional parts: $\frac{2}{5} - \frac{1}{5}$.

These are like fractions, so subtract the numerators:

$\frac{2 - 1}{5} = \frac{1}{5}$.

3. Combine the results from the whole and fractional parts:

$2 + \frac{1}{5} = \mathbf{2\frac{1}{5}}$.

---

Method 2: Converting to Improper Fractions

1. Convert the mixed fractions to improper fractions.

$4\frac{2}{5} = \frac{(4 \times 5) + 2}{5} = \frac{20 + 2}{5} = \frac{22}{5}$.

$2\frac{1}{5} = \frac{(2 \times 5) + 1}{5} = \frac{10 + 1}{5} = \frac{11}{5}$.

2. Now subtract the improper fractions: $\frac{22}{5} - \frac{11}{5}$.

3. Since these are like fractions, subtract the numerators and keep the common denominator:

$\frac{22 - 11}{5} = \frac{11}{5}$.

4. Convert the resulting improper fraction $\frac{11}{5}$ back to a mixed fraction.

Divide 11 by 5:

$\begin{array}{r} 2\phantom{)} \\ 5{\overline{\smash{\big)}\,11\phantom{)}}} \\ \underline{-~\phantom{(}10} \\ 1\phantom{)} \end{array}$

The quotient is 2, and the remainder is 1.

So, $\frac{11}{5} = \mathbf{2\frac{1}{5}}$.

---

Both methods give the same result. The difference is $\mathbf{2\frac{1}{5}}$.

Solution:

We need to calculate the difference $8\frac{1}{4} - 2\frac{5}{6}$.

Method 1: Converting to Improper Fractions

1. Convert both mixed fractions into improper fractions.

$8\frac{1}{4} = \frac{(8 \times 4) + 1}{4} = \frac{32 + 1}{4} = \frac{33}{4}$.

$2\frac{5}{6} = \frac{(2 \times 6) + 5}{6} = \frac{12 + 5}{6} = \frac{17}{6}$.

2. The expression becomes $\frac{33}{4} - \frac{17}{6}$.

3. Find the Least Common Multiple (LCM) of the denominators 4 and 6.

Multiples of 4: 4, 8, 12, ...

Multiples of 6: 6, 12, ...

LCM(4, 6) = 12.

4. Convert each improper fraction to an equivalent fraction with the denominator 12.

$\frac{33}{4} = \frac{33 \times 3}{4 \times 3} = \frac{99}{12}$.

$\frac{17}{6} = \frac{17 \times 2}{6 \times 2} = \frac{34}{12}$.

5. Subtract the resulting like fractions:

$\frac{99}{12} - \frac{34}{12} = \frac{99 - 34}{12} = \frac{65}{12}$.

6. Convert the improper fraction $\frac{65}{12}$ back to a mixed fraction.

Divide 65 by 12:

$\begin{array}{r} 5\phantom{)} \\ 12{\overline{\smash{\big)}\,65\phantom{)}}} \\ \underline{-~\phantom{(}60} \\ 5\phantom{)} \end{array}$

The quotient is 5, and the remainder is 5.

So, $\frac{65}{12} = \mathbf{5\frac{5}{12}}$.

---

Alternate Method: Subtracting by Borrowing

1. The expression is $8\frac{1}{4} - 2\frac{5}{6}$.

2. Find a common denominator for the fractional parts $\frac{1}{4}$ and $\frac{5}{6}$. LCM(4, 6) = 12.

$8\frac{1}{4} = 8\frac{1 \times 3}{4 \times 3} = 8\frac{3}{12}$.

$2\frac{5}{6} = 2\frac{5 \times 2}{6 \times 2} = 2\frac{10}{12}$.

3. The subtraction becomes $8\frac{3}{12} - 2\frac{10}{12}$.

4. Since the first fractional part ($\frac{3}{12}$) is smaller than the second ($\frac{10}{12}$), we need to borrow from the whole number part (8).

Borrow 1 from 8, leaving 7. The borrowed 1 is equal to $\frac{12}{12}$.

Add the borrowed $\frac{12}{12}$ to the existing fraction $\frac{3}{12}$: $8\frac{3}{12} = 7 + 1 + \frac{3}{12} = 7 + \frac{12}{12} + \frac{3}{12} = 7\frac{15}{12}$.

5. Now the subtraction is $7\frac{15}{12} - 2\frac{10}{12}$.

6. Subtract the whole parts: $7 - 2 = 5$.

7. Subtract the fractional parts: $\frac{15}{12} - \frac{10}{12} = \frac{15 - 10}{12} = \frac{5}{12}$.

8. Combine the results: $5 + \frac{5}{12} = \mathbf{5\frac{5}{12}}$.

---

The simplified result is $\mathbf{5\frac{5}{12}}$.

Solution:

To solve these, we add or subtract the fractions. If the fractions are unlike (different denominators), we first find the Least Common Multiple (LCM) of the denominators, convert the fractions to equivalent fractions with the LCM as the denominator, and then perform the operation.

---

(a) $\frac{2}{3} + \frac{1}{7}$

LCM(3, 7) = 21.

$\frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21}$

$\frac{1}{7} = \frac{1 \times 3}{7 \times 3} = \frac{3}{21}$

$\frac{14}{21} + \frac{3}{21} = \frac{14 + 3}{21} = \mathbf{\frac{17}{21}}$

---

(b) $\frac{3}{10} + \frac{7}{15}$

LCM(10, 15) = 30.

$\frac{3}{10} = \frac{3 \times 3}{10 \times 3} = \frac{9}{30}$

$\frac{7}{15} = \frac{7 \times 2}{15 \times 2} = \frac{14}{30}$

$\frac{9}{30} + \frac{14}{30} = \frac{9 + 14}{30} = \mathbf{\frac{23}{30}}$

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(c) $\frac{4}{9} + \frac{2}{7}$

LCM(9, 7) = 63.

$\frac{4}{9} = \frac{4 \times 7}{9 \times 7} = \frac{28}{63}$

$\frac{2}{7} = \frac{2 \times 9}{7 \times 9} = \frac{18}{63}$

$\frac{28}{63} + \frac{18}{63} = \frac{28 + 18}{63} = \mathbf{\frac{46}{63}}$

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(d) $\frac{5}{7} + \frac{1}{3}$

LCM(7, 3) = 21.

$\frac{5}{7} = \frac{5 \times 3}{7 \times 3} = \frac{15}{21}$

$\frac{1}{3} = \frac{1 \times 7}{3 \times 7} = \frac{7}{21}$

$\frac{15}{21} + \frac{7}{21} = \frac{15 + 7}{21} = \frac{22}{21}$

Convert to mixed fraction: $\frac{22}{21} = \mathbf{1\frac{1}{21}}$

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(e) $\frac{2}{5} + \frac{1}{6}$

LCM(5, 6) = 30.

$\frac{2}{5} = \frac{2 \times 6}{5 \times 6} = \frac{12}{30}$

$\frac{1}{6} = \frac{1 \times 5}{6 \times 5} = \frac{5}{30}$

$\frac{12}{30} + \frac{5}{30} = \frac{12 + 5}{30} = \mathbf{\frac{17}{30}}$

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(f) $\frac{4}{5} + \frac{2}{3}$

LCM(5, 3) = 15.

$\frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15}$

$\frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15}$

$\frac{12}{15} + \frac{10}{15} = \frac{12 + 10}{15} = \frac{22}{15}$

Convert to mixed fraction: $\frac{22}{15} = \mathbf{1\frac{7}{15}}$

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(g) $\frac{3}{4} - \frac{1}{3}$

LCM(4, 3) = 12.

$\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}$

$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$

$\frac{9}{12} - \frac{4}{12} = \frac{9 - 4}{12} = \mathbf{\frac{5}{12}}$

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(h) $\frac{5}{6} - \frac{1}{3}$

LCM(6, 3) = 6.

$\frac{5}{6}$ remains $\frac{5}{6}$.

$\frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6}$

$\frac{5}{6} - \frac{2}{6} = \frac{5 - 2}{6} = \frac{3}{6}$

Simplify: $\frac{3 \div 3}{6 \div 3} = \mathbf{\frac{1}{2}}$

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(i) $\frac{2}{3} + \frac{3}{4} + \frac{1}{2}$

LCM(3, 4, 2) = 12.

$\frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}$

$\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}$

$\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12}$

$\frac{8}{12} + \frac{9}{12} + \frac{6}{12} = \frac{8 + 9 + 6}{12} = \frac{23}{12}$

Convert to mixed fraction: $\frac{23}{12} = \mathbf{1\frac{11}{12}}$

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(j) $\frac{1}{2} + \frac{1}{3} + \frac{1}{6}$

LCM(2, 3, 6) = 6.

$\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}$

$\frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6}$

$\frac{1}{6}$ remains $\frac{1}{6}$.

$\frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = \mathbf{1}$

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(k) $1\frac{1}{3} + 3\frac{2}{3}$

Add whole parts: $1 + 3 = 4$.

Add fractional parts: $\frac{1}{3} + \frac{2}{3} = \frac{1+2}{3} = \frac{3}{3} = 1$.

Combine: $4 + 1 = \mathbf{5}$.

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(l) $4\frac{2}{3} + 3\frac{1}{4}$

Convert to improper fractions:

$4\frac{2}{3} = \frac{(4 \times 3) + 2}{3} = \frac{14}{3}$

$3\frac{1}{4} = \frac{(3 \times 4) + 1}{4} = \frac{13}{4}$

Add $\frac{14}{3} + \frac{13}{4}$. LCM(3, 4) = 12.

$\frac{14}{3} = \frac{14 \times 4}{3 \times 4} = \frac{56}{12}$

$\frac{13}{4} = \frac{13 \times 3}{4 \times 3} = \frac{39}{12}$

$\frac{56}{12} + \frac{39}{12} = \frac{56 + 39}{12} = \frac{95}{12}$

Convert to mixed fraction: $\frac{95}{12} = \mathbf{7\frac{11}{12}}$

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(m) $\frac{16}{5} - \frac{7}{5}$

These are like fractions.

$\frac{16 - 7}{5} = \frac{9}{5}$

Convert to mixed fraction: $\frac{9}{5} = \mathbf{1\frac{4}{5}}$

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(n) $\frac{4}{3} - \frac{1}{2}$

LCM(3, 2) = 6.

$\frac{4}{3} = \frac{4 \times 2}{3 \times 2} = \frac{8}{6}$

$\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}$

$\frac{8}{6} - \frac{3}{6} = \frac{8 - 3}{6} = \mathbf{\frac{5}{6}}$

Solution:

To find the total length of the ribbon bought by Sarita and Lalita, we need to add the lengths of ribbon each of them bought.

Length of ribbon bought by Sarita = $\frac{2}{5}$ metre.

Length of ribbon bought by Lalita = $\frac{3}{4}$ metre.

Total length = Length bought by Sarita + Length bought by Lalita

Total length = $\frac{2}{5} + \frac{3}{4}$ metres.

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The fractions $\frac{2}{5}$ and $\frac{3}{4}$ are unlike fractions because they have different denominators (5 and 4).

To add them, we first find the Least Common Multiple (LCM) of the denominators 5 and 4.

LCM(5, 4) = 20.

---

Now, we convert each fraction to an equivalent fraction with the denominator 20:

For $\frac{2}{5}$: Multiply numerator and denominator by 4 ($20 \div 5 = 4$).

$\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}$.

For $\frac{3}{4}$: Multiply numerator and denominator by 5 ($20 \div 4 = 5$).

$\frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}$.

---

Now, add the equivalent like fractions:

Total length = $\frac{8}{20} + \frac{15}{20}$

Total length = $\frac{8 + 15}{20} = \frac{23}{20}$ metres.

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The result $\frac{23}{20}$ is an improper fraction. We can convert it to a mixed fraction.

Divide 23 by 20:

$\begin{array}{r} 1\phantom{)} \\ 20{\overline{\smash{\big)}\,23\phantom{)}}} \\ \underline{-~\phantom{(}20} \\ 3\phantom{)} \end{array}$

The quotient is 1 and the remainder is 3.

So, $\frac{23}{20} = 1\frac{3}{20}$.

Therefore, the total length of the ribbon they bought is $\mathbf{1\frac{3}{20}}$ metres.

Solution:

To find the total amount of cake given to both Naina and Najma, we need to add the amount of cake each person received.

Amount of cake given to Naina = $1\frac{1}{2}$ pieces.

Amount of cake given to Najma = $1\frac{1}{3}$ pieces.

Total amount of cake = Amount for Naina + Amount for Najma

Total amount = $1\frac{1}{2} + 1\frac{1}{3}$.

---

We can solve this by adding the mixed fractions. Let's use the method of converting them to improper fractions first.

1. Convert the mixed fractions to improper fractions:

$1\frac{1}{2} = \frac{(1 \times 2) + 1}{2} = \frac{2 + 1}{2} = \frac{3}{2}$.

$1\frac{1}{3} = \frac{(1 \times 3) + 1}{3} = \frac{3 + 1}{3} = \frac{4}{3}$.

2. Now, add the improper fractions: $\frac{3}{2} + \frac{4}{3}$.

3. These are unlike fractions. Find the Least Common Multiple (LCM) of the denominators 2 and 3.

LCM(2, 3) = 6.

4. Convert the fractions to equivalent fractions with the denominator 6:

$\frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6}$.

$\frac{4}{3} = \frac{4 \times 2}{3 \times 2} = \frac{8}{6}$.

5. Add the like fractions:

$\frac{9}{6} + \frac{8}{6} = \frac{9 + 8}{6} = \frac{17}{6}$.

6. Convert the result $\frac{17}{6}$ back to a mixed fraction.

Divide 17 by 6:

$\begin{array}{r} 2\phantom{)} \\ 6{\overline{\smash{\big)}\,17\phantom{)}}} \\ \underline{-~\phantom{(}12} \\ 5\phantom{)} \end{array}$

The quotient is 2 and the remainder is 5.

So, $\frac{17}{6} = 2\frac{5}{6}$.

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Alternate Method: Adding Whole and Fractional Parts Separately

1. Add the whole parts: $1 + 1 = 2$.

2. Add the fractional parts: $\frac{1}{2} + \frac{1}{3}$.

LCM(2, 3) = 6.

$\frac{1}{2} = \frac{3}{6}$ and $\frac{1}{3} = \frac{2}{6}$.

$\frac{3}{6} + \frac{2}{6} = \frac{3+2}{6} = \frac{5}{6}$.

3. Combine the results: $2 + \frac{5}{6} = 2\frac{5}{6}$.

---

Therefore, the total amount of cake given to both of them was $\mathbf{2\frac{5}{6}}$ pieces.

Solution:

To find the missing fraction in each equation, we can rearrange the equation to isolate the unknown box.

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(a) $\boxed{?} - \frac{5}{8} = \frac{1}{4}$

Let the missing fraction be $x$. The equation is $x - \frac{5}{8} = \frac{1}{4}$.

To find $x$, we add $\frac{5}{8}$ to both sides:

$x = \frac{1}{4} + \frac{5}{8}$.

To add these fractions, we need a common denominator. The Least Common Multiple (LCM) of 4 and 8 is 8.

Convert $\frac{1}{4}$ to an equivalent fraction with denominator 8:

$\frac{1}{4} = \frac{1 \times 2}{4 \times 2} = \frac{2}{8}$.

Now add the like fractions:

$x = \frac{2}{8} + \frac{5}{8} = \frac{2 + 5}{8} = \frac{7}{8}$.

Therefore, the missing fraction is $\mathbf{\frac{7}{8}}$.

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(b) $\boxed{?} - \frac{1}{5} = \frac{1}{2}$

Let the missing fraction be $y$. The equation is $y - \frac{1}{5} = \frac{1}{2}$.

To find $y$, we add $\frac{1}{5}$ to both sides:

$y = \frac{1}{2} + \frac{1}{5}$.

Find the LCM of the denominators 2 and 5. LCM(2, 5) = 10.

Convert the fractions to equivalent fractions with denominator 10:

$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$.

$\frac{1}{5} = \frac{1 \times 2}{5 \times 2} = \frac{2}{10}$.

Now add the like fractions:

$y = \frac{5}{10} + \frac{2}{10} = \frac{5 + 2}{10} = \frac{7}{10}$.

Therefore, the missing fraction is $\mathbf{\frac{7}{10}}$.

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(c) $\frac{1}{2} - \boxed{?} = \frac{1}{6}$

Let the missing fraction be $z$. The equation is $\frac{1}{2} - z = \frac{1}{6}$.

To find $z$, we can rearrange the equation: $z = \frac{1}{2} - \frac{1}{6}$.

Find the LCM of the denominators 2 and 6. LCM(2, 6) = 6.

Convert $\frac{1}{2}$ to an equivalent fraction with denominator 6:

$\frac{1}{2} = \frac{1 \times 3}{2 \times 3} = \frac{3}{6}$.

Now subtract the like fractions:

$z = \frac{3}{6} - \frac{1}{6} = \frac{3 - 1}{6} = \frac{2}{6}$.

Simplify the result: $\frac{2}{6} = \frac{2 \div 2}{6 \div 2} = \frac{1}{3}$.

Therefore, the missing fraction is $\mathbf{\frac{1}{3}}$.

Solution:

We need to perform the additions (indicated by arrows pointing right) and subtractions (indicated by arrows pointing down) to complete the boxes.

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(a)

Calculations:

Row 1 addition ($\rightarrow$): $\frac{2}{3} + \frac{4}{3} = \frac{2+4}{3} = \frac{6}{3} = 2$.

Row 2 addition ($\rightarrow$): $\frac{1}{3} + \frac{2}{3} = \frac{1+2}{3} = \frac{3}{3} = 1$.

Column 1 subtraction ($\downarrow$): $\frac{2}{3} - \frac{1}{3} = \frac{2-1}{3} = \frac{1}{3}$.

Column 2 subtraction ($\downarrow$): $\frac{4}{3} - \frac{2}{3} = \frac{4-2}{3} = \frac{2}{3}$.

Bottom right box (check): $\frac{1}{3} + \frac{2}{3} = \frac{1+2}{3} = \frac{3}{3} = 1$. (Matches Row 1 sum - Row 2 sum: $2 - 1 = 1$).

Completed Box (a):

$\frac{2}{3}$	$\frac{4}{3}$	$2$
$\frac{1}{3}$	$\frac{2}{3}$	$1$
$\frac{1}{3}$	$\frac{2}{3}$	$1$

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(b)

Calculations:

Row 1 addition ($\rightarrow$): $\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{3+2}{6} = \frac{5}{6}$.

Row 2 addition ($\rightarrow$): $\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{4+3}{12} = \frac{7}{12}$.

Column 1 subtraction ($\downarrow$): $\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{3-2}{6} = \frac{1}{6}$.

Column 2 subtraction ($\downarrow$): $\frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{4-3}{12} = \frac{1}{12}$.

Bottom right box (check): $\frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{2+1}{12} = \frac{3}{12} = \frac{1}{4}$. (Matches Row 1 sum - Row 2 sum: $\frac{5}{6} - \frac{7}{12} = \frac{10}{12} - \frac{7}{12} = \frac{3}{12} = \frac{1}{4}$).

Completed Box (b):

$\frac{1}{2}$	$\frac{1}{3}$	$\frac{5}{6}$
$\frac{1}{3}$	$\frac{1}{4}$	$\frac{7}{12}$
$\frac{1}{6}$	$\frac{1}{12}$	$\frac{1}{4}$

Solution:

We are given the total length of the wire and the length of one of the pieces after it broke.

Total length of the wire = $\frac{7}{8}$ metre.

Length of one piece = $\frac{1}{4}$ metre.

To find the length of the other piece, we need to subtract the length of the known piece from the total length of the wire.

Length of the other piece = Total length - Length of one piece

Length of the other piece = $\frac{7}{8} - \frac{1}{4}$ metres.

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The fractions $\frac{7}{8}$ and $\frac{1}{4}$ are unlike fractions because they have different denominators (8 and 4).

To subtract them, we find the Least Common Multiple (LCM) of the denominators 8 and 4.

LCM(8, 4) = 8.

---

Now, we convert the fractions to equivalent fractions with the denominator 8. The fraction $\frac{7}{8}$ already has the denominator 8.

Convert $\frac{1}{4}$ to an equivalent fraction with denominator 8:

Multiply the numerator and denominator by 2 ($8 \div 4 = 2$).

$\frac{1}{4} = \frac{1 \times 2}{4 \times 2} = \frac{2}{8}$.

---

Now, perform the subtraction using the equivalent like fractions:

Length of the other piece = $\frac{7}{8} - \frac{2}{8}$

Subtract the numerators and keep the common denominator:

Length of the other piece = $\frac{7 - 2}{8} = \frac{5}{8}$ metres.

Therefore, the length of the other piece of wire is $\mathbf{\frac{5}{8}}$ metre.

Solution:

We are given the total distance from Nandini's house to her school and the distance she traveled by bus. We need to find the distance she walked.

Total distance from house to school = $\frac{9}{10}$ km.

Distance covered by bus = $\frac{1}{2}$ km.

The distance walked is the difference between the total distance and the distance covered by bus.

Distance walked = Total distance - Distance covered by bus

Distance walked = $\frac{9}{10} - \frac{1}{2}$ km.

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To subtract these unlike fractions, we need to find a common denominator. The Least Common Multiple (LCM) of the denominators 10 and 2 is 10.

---

Convert the fractions to equivalent fractions with the denominator 10.

The fraction $\frac{9}{10}$ already has the denominator 10.

Convert $\frac{1}{2}$ to an equivalent fraction with denominator 10:

Multiply the numerator and denominator by 5 ($10 \div 2 = 5$).

$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$.

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Now, perform the subtraction using the equivalent like fractions:

Distance walked = $\frac{9}{10} - \frac{5}{10}$.

Subtract the numerators and keep the common denominator:

Distance walked = $\frac{9 - 5}{10} = \frac{4}{10}$ km.

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Simplify the resulting fraction $\frac{4}{10}$. The Greatest Common Divisor (GCD) of 4 and 10 is 2.

Divide both numerator and denominator by 2:

Distance walked = $\frac{4 \div 2}{10 \div 2} = \frac{2}{5}$ km.

Therefore, Nandini walked $\mathbf{\frac{2}{5}}$ km.

Solution:

We need to compare the fractions representing how full each person's bookshelf is, and then find the difference between the larger and smaller fractions.

Fraction of Asha's shelf that is full = $\frac{5}{6}$.

Fraction of Samuel's shelf that is full = $\frac{2}{5}$.

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Comparison: Whose bookshelf is more full?

To compare $\frac{5}{6}$ and $\frac{2}{5}$, we need to find a common denominator. The Least Common Multiple (LCM) of the denominators 6 and 5 is 30.

Convert each fraction to an equivalent fraction with a denominator of 30:

Asha: $\frac{5}{6} = \frac{5 \times 5}{6 \times 5} = \frac{25}{30}$.

Samuel: $\frac{2}{5} = \frac{2 \times 6}{5 \times 6} = \frac{12}{30}$.

Now compare the numerators of the like fractions $\frac{25}{30}$ and $\frac{12}{30}$.

Since $25 > 12$, we have $\frac{25}{30} > \frac{12}{30}$.

This means $\frac{5}{6} > \frac{2}{5}$.

Therefore, Asha's bookshelf is more full.

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Difference: By what fraction?

To find out by what fraction Asha's shelf is more full, we subtract the smaller fraction (Samuel's) from the larger fraction (Asha's).

Difference = $\frac{5}{6} - \frac{2}{5}$.

Using the equivalent fractions with the common denominator 30:

Difference = $\frac{25}{30} - \frac{12}{30}$.

Subtract the numerators:

Difference = $\frac{25 - 12}{30} = \frac{13}{30}$.

Therefore, Asha's bookshelf is more full than Samuel's by a fraction of $\mathbf{\frac{13}{30}}$.

Solution:

We need to compare the time taken by Jaidev and Rahul and find the difference between their times.

Time taken by Jaidev = $2\frac{1}{5}$ minutes.

Time taken by Rahul = $\frac{7}{4}$ minutes.

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Comparison: Who takes less time?

First, let's convert Jaidev's time into an improper fraction:

$2\frac{1}{5} = \frac{(2 \times 5) + 1}{5} = \frac{10 + 1}{5} = \frac{11}{5}$ minutes.

Now we compare Jaidev's time ($\frac{11}{5}$ minutes) with Rahul's time ($\frac{7}{4}$ minutes).

To compare these unlike fractions, we find a common denominator. The Least Common Multiple (LCM) of the denominators 5 and 4 is 20.

Convert the fractions to equivalent fractions with a denominator of 20:

Jaidev's time: $\frac{11}{5} = \frac{11 \times 4}{5 \times 4} = \frac{44}{20}$ minutes.

Rahul's time: $\frac{7}{4} = \frac{7 \times 5}{4 \times 5} = \frac{35}{20}$ minutes.

Compare the numerators of the like fractions $\frac{44}{20}$ and $\frac{35}{20}$.

Since $35 < 44$, we have $\frac{35}{20} < \frac{44}{20}$.

This means $\frac{7}{4} < \frac{11}{5}$.

Therefore, Rahul takes less time than Jaidev.

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Difference: By what fraction?

To find out how much less time Rahul takes, we subtract Rahul's time (the smaller time) from Jaidev's time (the larger time).

Difference = Jaidev's time - Rahul's time

Difference = $\frac{11}{5} - \frac{7}{4}$ minutes.

Using the equivalent fractions with the common denominator 20:

Difference = $\frac{44}{20} - \frac{35}{20}$ minutes.

Subtract the numerators:

Difference = $\frac{44 - 35}{20} = \frac{9}{20}$ minutes.

Therefore, Rahul takes less time than Jaidev by $\mathbf{\frac{9}{20}}$ minutes.